Sastry CBSE

Electromagnetic Induction Important Questions for CBSE Class 12 Physics Electromagnetic Induction Laws

Previous Year Examination Questions

1 Mark Questions

1.The electric current flowing in a wire in the direction from B to A Find out the direction of the induced current in the metallic loop kept the wire as shown in the figure.

Ans.According to Lenz’s law, the direction of induced current will oppose the cause of its production. So, the current in loop will induce in such a way that it will support the current flowing in the wire i.e., in the same direction. So, the direction of current in the loop will be clockwise

2.Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same height above the ground. Which of the two wduld reach earlier and why? [Delhi 2014]
Ans.The glass bob will reach earlier on ground as acceleration due to gravity is independent of mass of the falling bodies. Being insulator, no induced current is developed in it due to the earth’s magnetic field.

3.A flexible wire of irregular shape, abed as shown in the figure, turns into a circular shape when placed in a region field which is directed normal to the plane of the loop away from the reader. Predict the direction of the induced current in the wire.[Foreign 2014]

Ans.The wire is expanding to from a circle, which means that force is acting outwards on each part of the wire because of the magnetic field (acting in the downwards direction). The direction of the induced current should be such that it will produce magnetic field in upward direction (towards the reader). Hence, the force on the wire will be towards inward direction, i.e. induced current is flowing in anti-clockwise direction in’the loop from cbadc.

4.In the given figure, a bar magnet is quickly moved towards a conducting loop having a capacitor. Predict the polarity of the plates A and B of the capacitor[All India 2014 C;2010;Hots]

Ans.

5.State Lenz’s law. A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer. [Delhi 2013]
Ans. Lenz’s law states that the induced emf of induced current is a circuit which always opposes the cause that produces it.Yes, emf will be induced in the rod as there is change in magnetic flux.When a metallic rod held horizontally along East-West direction, is allowed to fall freely under gravity i.e. fall from North to South, the intensity of magnetic lines of the earth’s magnetic field changes through it, i.e. the magnetic flux changes and hence the induced emf in it

6.State the Faraday’s law of electromagnetic induction. [Foreign 2009,2012]
Ans.

7.The closed loop PQRS of wire is moved into a uniform magnetic field at right angles to the plaue of the paper as shown in the figure. Predict the direction of the induced current in the loop.

Ans.Since, magnetic flux increasing when the loop moves into uniform magnetic field. So the induced current should oppose this increase. Thus, flow will be from QPSRQ. i.e. anti-clockwise

8.Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing?

Ans.Current in the wire is steadily decreasing, so the induced current in rings 1 and 2 will flow in such a way that it opposes the decrease of current

So, it will flow in same direction. Now, from the figure, it is clear that the direction of induced current in (i) ring 1 is clockwise.
(ii) ring 2 is anti-clockwise

9.A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the directions of induced current in each coil.

Ans.

2 Marks Questions

10.Predict the polarity of the capacitor in the situation described by adjoining as shown in figure. Explain the reason too.

Ans.According to figure shown in the question, induced current is in anti-clockwise, when seen from left hand side and its direction is in clockwise when seen from right hand side. Thus, direction of induced     current  is in clockwise sense.This implies that plate A of the capacitor is at the higher potential than plate B, i.e. B is as negative plate while  A  is  as  positive plate.

11.A metallic rod of length L is rotated with angular frequency of co with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. [Delhi 2012]
Ans.

12.A current is induced in coil C₁ due to the motion of current carrying coil C₂.

(i)Write any two ways by which a large deflection can be obtained in the galvanometer G
(ii)Suggest an alternative device to demonstrate the induced current in place of a galvanometer. [Delhi 2011]
Ans.

(a)by moving quickly, the coil C₂ towards C1, or by moving quickly the coil C₂ away from C,.
(b)by switching off and on the key.
(ii) Alternating device in place of galvanometer can be LED or bulb.

13.A magnet is quickly moved in the direction indicated by an arrow between two coils Q and C₂ as shown in the figure.

What will be the direction of induced current in each coil as seen from the magnet? Justify your answer.   [Foreign 2011]
Ans.As N-pole of the magnet is moving away from the coil C1 therefore the coil will behave as South-pole and opposes the motion of the magnet. This implies the current in the coil is anti-clockwise. Also in coil C₂, it behaves as S-pole in order to repel the coming magnet. This implies current in the coil is anti-ciockwise.

14.Two identical loops, one of copper and the other of aluminium are rotated with the same angular speed in the same magnetic field. Compare
(i)the induced emf and
(ii)the current produced in the two coils. Justify your answer.[All India 2010]
Ans.(i)The induced emf in both the loops will be same as areas of the loop and time periods are same as they are identical and rotated with same angular speed.
(ii)The current induces in Cu coil is more than Al coil as Cu coil have got lesser resistance and

15.(i)When primary coil P is moved towards secondary coil S (as shown in the figure below), the galvanometer shows momentary deflection. What can be done to have larger deflection in the galvanometer with the same battery?

Ans.The coil P should be moved quickly towards or away from the coil 5.The laws involved here are Faraday’s law of electromagnetic induction.

16.A coil O is connected to low voltage bulb B and placed near another coil P as shown in the figure. Give reasons to explain the following observations.
(i)The bulb Blights.
(ii)Bulb gets dimmer. If the coil Q is moved towards left.

Ans.

17.State the law that gives the polarity of the induced emf.     [All India 2009]
Ans. Lenz’s law gives the polarity of induced emf. The induced emf (or current) always opposes the cause that produces it.
Lenz’s law The induced emf of induced current is a circuit which always opposes the cause that produces it.

When North pole of the bar magnet approaches towards the loop, the induced current in coil is anti-clockwise (forming North pole of current loop) when viewed from magnetic side. The face of coil being North pole opposes the arrival of North pole of magnet. Hence, opposes the cause that produces it.   Also, certain amount of work has to be done by external agency to bring the North pole near the coil in against of force of repulsion applied by induced current loop. The work done by external agency appears in the form of electrical energy. So, Lenz’s law is a consequence of the principle of law of conservation of energy

18.State Lenz’s law. Give one example to illustrate this law. The Lenz’s law is a consequence of the principle of conservation of energy. Justify this statement.   [All India 2009 ]
Ans.Lenz’s law The induced emf of induced current is a circuit which always opposes the cause that produces it.

When North pole of the bar magnet approaches towards the loop, the induced current in coil is anti-clockwise (forming North pole of current loop) when viewed from magnetic side. The face of coil being North pole opposes the arrival of North pole of magnet. Hence, opposes the cause that produces it.     Also, certain amount of work has to be done by external agency to bring the North pole near the coil in against of force of repulsion applied by induced current loop. The work done by external agency appears in the form of electrical energy. So, Lenz’s law is a consequence of the principle of law of conservation of energy

3 Marks Questions
19.(i) A rod of length / is moved horizontally with a uniform velocity v in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.
(ii) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain. [All India 2014]
Ans.(i)Consider a straight conductor moving with velocity v and U shaped conductor placed in perpendicular magnetic field as shown in the figure.

Ans.

Component of velocity v parallel to magnetic field, will make the particle move along the field. Perpendicular component of velocity v, will cause the particle to move along a circular path in the plane perpendicular to the magnetic field. Hence, the particle will travel the helix path.

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23.A conducting rod PQ, of length /, connected to a resistor R, is moved at a uniform speed v, normal to a uniform magnetic field B, as shown in the figure

(i)Deduce the expression for the emf induced in the conductor.
(ii)Find the force required to move the rod in the magnetic field.
(iii)Mark the direction of induced current in the conductor.[Foreign 2008]
Ans.

24.Figure shows a rectangular conducting loop PQRS in which arm RS of length / is movable. The loop is kept in a uniform magnetic field B directed downward perpendicular to the plane of the loop. The arm RS is moved with a uniform speed v.

Deduce an expression for
(a)the emf induced across the arm RS
(b)the external force required to move the arm and
(c)the power dissipated as heat.[All India 2009]
Ans.

5 Marks Questions

25.(i) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it. (ii) The current flowing through an inductor of self-inductance L is continuously increasing. Plot a graph showing the variation of
(a)magnetic flux versusthe current
(b)induced emf versus dl/dt
(c)magnetic potential energy stored versusthe current.[Delhi 2014]
Ans.(i)The coil P should be moved quickly towards or away from the coil 5.The laws involved here are Faraday’s law of electromagnetic induction.

26.A metallic rod of length land resistance Ris rotated with a frequency v, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius /, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field ft parallel to the axis is present everywhere.
(a) Derive the expression for the induced emf and the current in the rod.
(b)Due to the presence of the . current in the rod and of the magnetic field, find the expression for the magnitude and direction of the force acting  on this rod.
(c)Hence, obtain the expression for the power required to rotate the rod.[All India 2014 C]
Ans.

Ans.



Important Questions for Class 12 PhysicsClass 12 PhysicsNCERT Solutions Home Page

Magnetism and Matter Important Questions for CBSE Class 12 Physics Earth’s Magnetic Field and Magnetic Material

Previous Years’ Examination Questions

1 Mark Questions

Ans. The nature of magnetic material is diamagnetic.

2.What are permanent magnets? Give one example.  [Delhi 2013]

Ans.Permanent magnets are those magnets which have high retentivity and coercivity The magnetisation of permanent magnet is not easily destroyed even if it is handled roughly or exposed in stray reverse magnetic field, e.g. steel.

3.Where on the surface of  the earth’s vertical component of earth’s magnetic field zero?  [Delhi 2013C, All india 2011]

Ans. At equator, vertical component will be zero.

4.The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of the earth’s magnetic field at equator? [Delhi 2012]

Ans.

5.What is the angle of dip at a place where the horizontal and vertical components of the earth’s magnetic field are equal?         [Foreign 2012]

Ans.

6.A magnetic needle free to rotate m a vertical plane orients itself vertically at a certain place on the earth What are the values of

• horizontal component df the earth’s magnetic field and
• angle of dip at this place? [Foreign 2012]

Ans.  (i) The coil is free to move in vertical plane, it means that there is no componerr A the earth’s magnetic field in horizontal direction, so the horizontal component of the earth’s magnetic field is 0.

(ii) The angle of dip is 90°.

7.Where on the surface of earth is the angle of dip 90°?                      [All India 2011]

Ans. At poles, the angle of dip is 90°.

8.The permeability of a magnetic material is 0.9983. Name the type of magnetic material, it represents. [Delhi 2011]

Ans.

9.The susceptibility of a magnetic  material is 1.9 x 10^-5. Name the of magnetic material, it represents. [Delhi 2011]

Ans.  The small and positive susceptibility of 1.9 x 10-⁵ represents paramagnetic substance.

Ans.  Negative susceptibility diamagnetic substance.

Ans. 

12. What is the Characteristic property of a diamagnetic material? [Foreign 2010]

Ans. Diamagnetic material acquires magnetisation in the opposite direction of the magnetic field when they are placed in an external magnetic field

13. Define the term magnetic declination. [All India 2009c]

Ans. Magnetic declination The angle between geographical meridian and magnetic meridian at any place of the earth is known as magnetic declination (a) at that place of the earth.

2 Marks Questions

14.Show diagrammatrically the behaviour of magnetic field lines in the presence of

• paramagnetic and
• diamagnetic substances. How does one explain this distinguishing feature? [All ihdia 2014]

Ans.  Magnetic permeability of paramagnetic is more than air, so it allows more lines to pass through it while permeability of diamagnetic is less than air, so it does not allow lines to pass through it.

For figure, refer ans. 34.

This difference can be explained as diamagnetic substances repel or expel the magnetic field lines while paramagnetic substance attract the magnetic field lines

15.Out of the two magnetic materials, A has relative permeability slightly greater than unity while B has less than unity. Identify the nature of the materials A and B. Will their susceptibilities   be positive  r negative?                             [Delhi 2014]

Ans. 

16. Give two points to distinguish between a paramagnetic and diamagnetic substance.

Ans.

17. (i) How is an electromagnet different from a permanent magnet?

(ii) Write two properties of a material which makes it suitable for making electromagnet. [All India 2014C]

Ans.  An electromagnet consists of a core made of a ferromagnetic material placed inside a solenoid. It behaves like a strong magnet when current flows through the solenoid and effectively loses its magnetism when the current is switched off.

(i) A permanent magnet is also made up of a ferromagnetic material but it retains its magnetism at room temperature for a long time after being magnetised one.

(ii) Properties of material are as below:

• High permeability
•  Low retentivity
• Low coercivity

18. The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its two properties. [Delhi 2012]

Ans. Ferromagnetic substance as these substances have very high magnetic permeability.

Properties (i) High retentivity

(ii) High susceptibility

19.(i) How does a diamagnetic material behave when it is cooled to very low temperature?

(ii) Why does a paramagnetic sample display greater magnetisation when cooled? Explain. [Delhi 2012]

Ans. (i) As, the resistance (electrical of metal decreases with decrease in temperature.

But for diamagnetic substances, the variation of susceptibility is very small, i.e. diamagnetic materials are unaffected by the change in temperature (except bismuth).                                               (1)

(ii) Paramagnetic materials when cooled due to thermal agitation tendency alignment of magnetic dipoles decreases. Hence, they shows greater magnetisation.

20.A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its North tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place.  [Delhi 2011]

Ans.

21.Distinguish between diamagnetic and ferromagnetic materials in terms of

• susceptibility and
• their behaviour in a non-uniform magnetic field.           [All India 2011]

Ans.

22.(i) Write two characteristics of a material used for making permanent magnets?

(ii) Why is core of an electromagnet made of ferromagnetic materials? [Delhi 2010]

Ans.

23.The horizontal component of the earth’s magnetic field at a place is V3 times its vertical component there. Find the value of the angle of dip at that place. What is the ratio of the horizontal component to the total magnetic field of the earth at that place?          [All India 2010c]

Ans.

24.The horizontal component of the earth’s magnetic field at a place equals to its vertical component there. Find the value of the angle of dip at that place. What is the ratio of the horizontal component to the total magnetic field of the earth at that place? [HOTS, All India 2010C]

Ans.

Ans.

Ans.

Ans.

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Ans.

Ans.

31.Define magnetic susceptibility Of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify? [Delhi 2008]

Ans.

32.Draw a plot showing the variation of intensity of magnetisation with the applied magnetic field intensity for bismuth. Under what condition cfbes a diamagnetic material exhibit perfect conductivity and perfect diamagnetism?     [Foreign 2008]

Ans.

3 Marks Questions

33.A wheel with 8 metallic spokes each 50 cm long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the each’s magnetic field. The earth’s magnetic field at the place is 0.4 G and the angle of dip is 60°. Calcu] ate the emf induced between the 3txle and the rim of wheel. How will the value of emf be affected if the number of spokes were increased? [All India 2013]

Ans.

Ans.

35.(i) What happens when  a diamagnetic substance is placed in a varying magnetic field?

(ii) Name the properties of a magnetic material that makes it suitable for making (a a permanent magnet and (b) a core of an electromagnet. [Foreign 2009]

Ans.

36.(i) How does angle of dip change as line goes from magnetic pole to magnetic equator of the earth?

(ii) A uniform magnetic field gets modified as shown in the figure below, when two specimens X and Y are placed in it. Identify whether specimens X and Y are diamagnetic, paramagnetic or ferromagnetic. [Foreign 2009]

Ans.

37.When two materials are placed in an external magnetic field, the behaviour of magnetic field lines is as shown in the figure. Identify the magnetic nature of each of these two materials.    [Delhi 2009c]

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38.Name the three elements required to specify the earth’s magnetic field at a given place. Draw a labelled diagram to define these elements. Explain briefly how these elements are determining to find out the magnetic field at a given place on the surface of the earth. [Delhi 2008]

Ans.

39

• Write the range of relative  magnetic permeability of the materials .
• Draw the pattern of magnetic field lines when the  materials are placed in external magnetic field. [Delhi 2008]

Ans.

40.Name and define the elements of the earth’s magnetic field other than the horizontal component of the earth’s magnetic field. Why do we say that at the place like Delhi and Mumbai, a magnetic needle shows the true North direction quite accurately as compared to other places in India? [All India 20011c]

Ans.

5 Marks Questions

41. (i) A small compass needle of magnetic moment M is free to turn about an axis perpendicular to the direction of uniform magnetic field B. The moment of inertia of the needle about the axis is I. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence, deduce , the expression for its time period, (ii) A compass needle free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. Find out the values of (a) horizontal component of the earth’s magnetic field and (b) angle of dip at the place.        [Delhi 2013]

Ans.



Important Questions for Class 12 PhysicsClass 12 PhysicsNCERT Solutions Home Page

Moving Charges and Magnetism Important Questions for CBSE Class 12 Physics Magnetic Field Laws and their Applications

1.The space in the surroundings of a magnet or a current-carrying conductor in which its magnetic influence can be experienced is called magnetic field. Its SI unit is Tesla (T).

2.Oersted experimentally demonstrated that the current-carrying conductor produced magnetic field around it.

Previous Year Examination Questions

1 Mark Question

1.Draw the magnetic field lines due to a current carrying loop.  [Delhi 2013 C]
Ans.

2 Marks Questions

2.Considering the case of a parallel plate capacitor being charged, show how one is required to generalise Ampere’s circuital law to include the term due to displacement current.  [All India 2014,2011]
Ans.

3.(i) State Biot-Savart’s law in vector form expressing the magnetic field due to an element dl carrying current 7 at a distance r from the element.
(ii) Write the expression for the magnitude of the magnetic field at the centre of a circular loop of radius r carrying a steady current 7. Draw the field lines due to the current loop.[All India 2014C]
Ans.(i)

(ii)

4.Define one tesla using the expression for the magnetic force acting on a particle of charge q moving with velocity v in a magnetic field B.  [Foreign 2014]
Ans.

5.Two identical circular loops P and Q, each of radius r and carrying equal currents are kept in the parallel planes having a common axis passing through O. The direction of current in P is clockwise and in Q is anti-clockwise as seen from O which is equidistant from the loops P and Q. Find the magnitude of the net magnetic field at O.

Ans.

6.A circular coil of closely wound N turns and radius r carries a current 7. Write the expressions for the following:
(i)The magnetic field at its centre.
(ii)The magnetic moment of this COil. [All India 2012]
Ans.

7.A particle of charge q and mass m is moving with velocity It is subjected to a uniform magnetic field B directed perpendicular to its velocity. Show that it describes a circular path. Write the expression for its radius.[Foreign 2012]
Ans.

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9.A long solenoid of length L having N turns carries a current I.Deduce the expression for the magnetic field in the interior of the solenoid. [All India 2008; 2011C]
Ans.

where number of turns per unit length. This is required expression for magnetic field inside the long current carrying solenoid.

10.Obtain with the help of a necessary diagram, the expression for the magnetic field in the interior of a toroid carrying current. [HOTS; All India 2011C]
Ans.

11.A straight wire of length L is bent into a semi-circular loop. Use Biot-Savart’s law to deduce an expression for the magnetic field at its centre due to the current 7 passing through it. [Delhi 2011c]
Ans.

12.State Ampere’s circuital law. Show through an example, how this law enables an easy evaluation of the magnetic field when there is a symmetry in the system? [All India 2010]
Ans.

13.State Biot-Savart’s law. A current I flows in a conductor placed perpendicular to the plane of the paper. Indicate the direction of the magnetic field due to a small element dl at a point P situated at a distance r from the element as shown in the figure.[Delhi 2009]

Ans.

14.A wire of length L is bent round in the form of a coil having N turns of same radius. If a steady current I flows through it in clockwise direction, then find the magnitude and direction of the magnetic field produced at its centre.[Foreign 2009]
Ans.

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16.Using Ampere’s circuital law,obtain an expression for the magnetic field along the axis of a current carrying solenoid of length / and having N number of turns.  [All India 2008]
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17.A circular coil of radius R carries a current /. Write the expression for the magnetic field due to this coil at its centre. Find out the direction Of the field. [All India 2008 C]
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3 Marks Questions

18.(i) State Ampere’s circuital law expressing it in the integral form, (ii) Two long co-axial insulated solenoids and S₂ of equal length are wound one over the other as shown in the figure.A steady current / flows through the inner solenoid S_x to the other end B which is connected to the outer solenoid through which the some current / flows in the opposite direction so, as to come out at end A. If n_x and n₂ are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point
(a)inside on the axis and
(b)outside the combined system

Ans.

19.(i) How is a toroid different from a solenoid?
(ii) Use Ampere’s circuital law to obtain the magnetic field inside a toroid.
(iii)Show that in an ideal toroid the ¹ magnetic field (a) inside the toroid and (b) outside the toroid at any point in the open space is zero.   [All India 2014 C]
Ans.(i) A toroid can be viewed as a solenoid which has been bent into circular shape to close on itself.
(ii)

(iii)For the evaluation of magnetic field for a symmetrical system, we can consider the example of a current carrying solenoid. Now,

where number of turns per unit length. This is required expression for magnetic field inside the long current carrying solenoid.

Ans.

Ans.

22.A solenoid of length 1.0 m has a radius of 1 cm and has a total of 1000 turns wound on it. It carries a current of 5 A. Calculate the magnitude of the axial magnetic field inside the solenoid. If an electron was to move with a speed of 10⁴ m/s along the axis of this current carrying solenoid, what would be the force experienced by this electron?  [Delhi 2008 C]
Ans.

23.A long straight wire of a circular cross-section of radius a carries a steady current /. The current is uniformly distributed across the cross-section of the wire. Use Ampere’s circuital law to show that the magnetic field due to this wire in the region inside the wire increases in direct proportion to the distance of the field point from the axis of the wire. Write the value of this magnetic field on the surface of the wire.[All India 2008 C]
Ans.

5 Marks Questions

24.Two very small identical circular loop(1) and (2) carrying equal current I are placed vertically (with respect to the plane of the paper) with their geometrical axes perpendicular to each other as shown in the figure. Find the magnitude and direction of the net magnetic field produced at the point O.[Delhi 2014]

Ans.

Ans..

(i)Let us consider a circular loop of radius a with centre C. Let the plane of the coil be perpendicular to the plane of the paper and current / be flowing in the direction shown. Suppose P is any point on the axis at direction r from the centre.

Now, consider a current element Idl on top L, where current comes out of paper normally, whereas at bottom M enters into the plane paper normally

26.State Biot-Savart’s law and give the mathematical expression for it. Use law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis. How does a circular loop carrying current behave as a magnet?[Delhi 2011]
Ans.For Biot-Savart’s law

For the magnetic field due to a circular coil carrying current at a point along its axis

(i)Let us consider a circular loop of radius a with centre C. Let the plane of the coil be perpendicular to the plane of the paper and current / be flowing in the direction shown. Suppose P is any point on the axis at direction r from the centre.

Now, consider a current element Idl on top L, where current comes out of paper normally, whereas at bottom M enters into the plane paper normally

As current carrying loop has the magnetic held lines around it which exists a force on a moving charge. Thus, it behaves as a magnet with two mutually opposite   poles

The anti-clockwise flow of current behaves like a North pole, whereas clockwise flow as South pole. Hence, loop behaves as a magnet

27.(i) Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis.
(ii) In what respect, is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.
(iii) How is the magnetic field inside a given solenoid made strong?[All India 2011]
Ans.(i)

where, n = number of turns per unit length. This is required expression for magnetic field inside the long current carrying solenoid.

(ii) Solenoid is a hollow circular ring having large number of turns of insulated copper wire on it.
Therefore, we can assume that toroid is a bent solenoid to close on itself.

The magnetic fields due to solenoid and toroid is given in figures below

Magnetic field inside the solenoid is a uniform, strong and along its axis also field lines are all most parallel while inside the toroid field line makes closed path.
(iii) The magnetic field in the solenoid can be increased by inserting a soft iron core inside it.

28.(i) State Ampere’s circuital law.
(ii) Use it to derive an expression for magnetic field inside along the axis of an air cored solenoid.
(iii) Sketch the magnetic field lines for a finite solenoid. How are these field lines different from the electric field lines from an electric dipole?   [Foreign 2010]
Ans.(i)For statement of Ampere’s circuital law


(ii)

where, n = number of turns per unit length. This is required expression for magnetic field inside the long current carrying solenoid.

29.(i)Using Biot-Savart’s law, deduce an expression for the magnetic field on the axis of a circular current carrying loop.
(ii) Draw the magnetic field lines due to a current carrying loop.
(iii) A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in the figure. What is the magnetic field B at O due to
(a)straight segments,
(b)the semi-circular arc?[Foreign 2010]

Ans.(i)Let us consider a circular loop of radius a with centre C. Let the plane of the coil be perpendicular to the plane of the paper and current / be flowing in the direction shown. Suppose P is any point on the axis at direction r from the centre.

Now, consider a current element Idl on top L, where current comes out of paper normally, whereas at bottom M enters into the plane paper normally

30.(i) State Ampere’s circuital law. Show through an example, how this law enables an easy evaluation of this magnetic field when there is a symmetry in the system?
(ii) What does a toroid consist of? Show that for an ideal toroid of closely wound turns, the magnetic field.
(a)inside the toroid is constant.
(b)in the open space inside an exterior to the toroid is zero.[All India 2010 C]
Ans.(i) For statement of Ampere’s law

For the evaluation of magnetic field for a symmetrical system, we can consider the example of a current carrying solenoid. Now,

where, n = number of turns per unit length. This is required expression for magnetic field inside the long current carrying solenoid.


Important Questions for Class 12 PhysicsClass 12 PhysicsNCERT Solutions Home Page

Magnetism and Matter Important Questions for CBSE Class 12 Physics Magnetic Dipole and Magnetic Field Lines

1.The magnetic dipole moment of a magnetic dipole is given by M = mx 21

where, m is pole strength and 21 is dipole length directed from The SI unit of magnetic dipole moment is A-m² or J/T. It is a vector quantity and its direction is from South pole to North pole.

2.Coulomb’s Law in Magnetism Magnitude of force acting between two magnetic poles is given by

where, m₁ and m₂ are magnetic strength of poles and k is magnetic force constant. Its SI unit is A-m.

3.Magnetic Field Lines These are imaginary lines which give pictorial representation for the magnetic field inside and around the magnet.

Their properties are given as below:

• These lines form continuous closed loops.
• The tangent to the field line gives direction of the field at that point.
• Larger the density of the lines, stronger will be the magnetic field.
• These lines do not intersect one another

5.Torque on a bar magnet in a uniform magnetic field is

6.Potential energy of a magnetic dipole in a magnetic field is given by


8.Current loop behaves like a magnetic dipole whose dipole moment is given by
M=IA
The direction of dipole moment can be obtained by right hand thumb rule. Its. SI unit is A-m².

9.Magnetic dipole moment of a revolving electron is given by

10.Interaction between two magnetic dipoles is

11. Oscillation of a Freely Suspended Magnet The oscillations of a freely suspended magnet (magnetic dipole) in a uniform magnetic field are SHM.

where, I – moment of inertia of the magnet, M = magnetic moment and B – magnetic field intensity.

12.Bar Magnet as an Equivalent Solenoid The expression of magnetic field at distance r from centre is given by

This expression is equivalent to that of bar magnet.

13.The Electrostatic Analog

The table given below summarises the analogy between electric and magnetic dipoles

14.Magnetism and Gauss’ Law

Previous Years’ Examination Questions

2 Marks Questions

1.Draw the magnetic field lines due to a current passing through a long solenoid. Use Ampere’s circuital law, to obtain the expression for the magnetic field due to the current I in a long solenoid having n number of turns per unit length. [Delhi 2014c]

Let a and b be two long straight parallel conductors. I_a and I_b are the current flowing through them and separated by a distance d. Magnetic field induction at a point P on a conductor b due to current I_a passing through a is

 

2.(i) Two long straight parallel conductors a and b carrying steady currents I_a and I_b respectively are separated by a distance d. Write the magnitude and direction, what is the nature and magnitude of the force between the two conductors?

(ii) Show with the help of a diagram, how the force between the two conductors would change when the currents in them flow in the opposite directions. [Foreign 2014]

Now, let the direction of current in conductor b be reversed. The magnetic field B₂ at point P due to current I_a flowing through a will be downwards. Similarly, the magnetic field B_: at point Q due to current I_b passing through b will also be downward as shown. The force on a will be, therefore towards the left. Also, the force on b will be towards the right. Hence, the two conductors will repel each other as shown.

3.A circular coil of N turns and radius R carries a current I. It is unwound and rewound to make another coil of radius R/2, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil.   [All India 2012]

4.A circular coil of N turns and diameter d carries a current I. It is unwound and rewound to make another coil of diameter 2d, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil.      [All India 2012]

5.Explain the following:

• Why do magnetic lines of force form continuous closed loops?
• Why are the field lines repelled (expelled) when a diamagnetic material is placed in an external uniform magnetic field? [Foreign 2011]

6.A small compass needle of magnetic moment M and moment of inertia I is free to oscillate in a magnetic field It is slightly disturbed from its equilibrium position and then released. Show that it executes simple harmonic motion. Hence, write the expression for its time period. [HOTS, Delhi 2011C]

7.How does a circular loop carrying current behaves as a magnet? [Delhi 2011]

Ans.The current round in the face of the coil is in anti-clockwise direction, then this behaves like a North pole, whereas when it viewed from other sc le, then current round in it is in clockwise direction necessarily forming South pole of magnet.

Hence, current loop have both magnetic poles and therefore, behaves like a magnetic dipole

8.Deduce the expression for the magnetic dipole moment of an electron orbiting around the central nucleus.         [All India 2010, Foreign 2009]

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Current Electricity Important Questions for CBSE Class 12 Physics Kirchhoff’s Laws and Electric Devices

1.Kirchhoff has given two rules based on conservation of electric charge and of energy, these are known as Kirchhoff’s laws

2.Kirchhoff’s First Law or Junction Rule The algebraic sum of electric currents at any junction of electric circuit is equal to zero i.e. the sum of current entering into a junction is equal to the sum of current leaving the junction i.e.

Junction law supports law of conservation of charge because this is a point in a circuit which cannot act as a source or sink of charge(s).
Kirchhoff’s Second Law or Voltage Rule In any closed mesh of electrical circuit, the algebraic sum of emf’s of cell and the product of currents and resistance is always equal to zero i.e.

Kirchhoff’s second law supports the law of conservation of energy. Because the net change in the energy of a charge, after the charge complete a closed path must be zero
NOTE Sign convention for Kirchhoffs second law

4.Wheatstone Bridge It is an arrangement of four resistance connected to form the arms of quadrilateral A battery with key and galvanometer are connected along its two diagonals respectively.
where, P,Q = ratio arms, R = known resistance
S = unknown resistance
The bridge is said to be balanced, when
(i) Vb = V_d
(ii) There is no flow of current through galvanometer.
The Wheatstone bridge is said to be sensitive, if it gives ample deflection in the galvanometer even on slight change of resistance.
For sensitivity of galvanometer, the magnitude of four resistances P, Q, R, S should be of same order.

5.Meter Bridge

It is an electrical device used to determine the resistance and hence, specific resistance of material of given wire/conductor.
It is based on the principle of balanced Wheatstone bridge.
For uniform wire, resistance of wire Length of conductor At balanced situation of bridge

Previous Year Examination Questions

1 Mark Questions

1.A heating element is marked 210 V, 630 What is the value of the current drawn by the element when connected to a 210 V DC Source? [Delhi 2013]
Ans.

2.In an experiment on meter bridge, if the balancing length AC is X, what would be its value, when the radius of the meter bridge wire AB is doubled? Justify your answer.[All India 2011 C]
Ans.

3.In a meter bridge, two unknown resistances R and S when connected in the two gaps, give a null point at 40 cm from one end. What is the ratio of R and S?[Delhi 2010]
Ans.

4.Use Kirchhoff’s rules to determine the value of the current I_x flowing in the circuit shown in the figure.

Ans.

5.In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B

Ans.

6.In the meter bridge experiment, balance point was observed at J with AJ =l.
(i)The values of R and X were doubled and then interchanged. What would be the new position of balance point?
(ii)If the galvanometer and battery are interchanged at the balanced position, how will the balance point get affected?

Ans.

7.Using Kirchhoff’s rules in the given circuit, determine
(i)the voltage drop across the unknown resistor R and
(ii)the current I₂ in the arm EF

Ans.

8.Calculate the current drawn from the battery in the given network

Ans.

9. For the circuit diagram of a Wheatstone bridge shown in the figure, use Kirchhoff’s laws to obtain its balance condition.

Ans.

10.Obtain the formula for the power loss (i.e. power dissipated) in a conductor of resistance R, carrying a current.[Delhi 2009 C]
Ans.

11.The adjoining graph shows the variation of terminal potently difference V, across a combination o three cells in series to a resistor vers¹ the current I.

(i)Calculate the emf of each cell.
(ii)For what current I, will the power dissipation of the circuit be maximum? [All India 2008]
Ans.

3 Marks Questions

12.Answer the following
(i)Why are the connections between the resistor in a meter bridge made of thick copper strips?
(ii)Why is it generally preferred to obtain the balance point in the middle of the meter bridge wire?
(iii)Which material is used for the meter bridge wire and why?[All India 2014]
Ans.(i)The connections between the resistors in a meter bridge are made of thick copper strips because of their negligible resistance.

(ii) It generally preferred to obtain the balance point in the middle of the meter bridge wire because meter bridge is most sensitive when all four resistances are of same order

(iii)Alloy, magnanin or constantum are used for making meter bridge wire due to low temperature coefficient of resistance and high resistivity

Ans.

14.Define the current sensitivity of a galvanometer. Write its SI unit.Figure shows two circuits each having a galvanometer and a battery of 3 V.When the galvanometer in each arrangement do not show any deflection, obtain the ratio R1 /R₂.

Ans.

Ans.

16.Calculate the value of the resistance R in the circuit shown in the figure, so that the current in the circuit is 0.2A. What would be the potential difference between points A and B ?

Ans.

17.Two  heating elements of resistances R1 and R₂ when operated at a constant supply of voltage V, consume powers P1 and P₂, respectively. Deduce the expressions for the power of their combination when they are, in turn, connected in
(i)series and
(ii)parallel across their same voltage supply.[All India 2011]
Ans.

Ans.

19. State Kirchhoff’s rules. Use these rules to write the expressions for the currents I_lt I₂ and I₃ in the circuit diagram shown in figure below.

Ans.

20.State Kirchhoff’s rules. Apply Kirchhoff’s rules to the loops ACBPA  and  ACBQA   to write the expressions for the currents I1,I₂ and I₃ in the network.

Ans.

21.State Kirchhoff’s rules. Apply these rules to the loops PRSP and PRQP to write the expressions for the currents I1, I₂ and I₃ in given circuit.

Ans.

Ans.

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24.In a meter bridge, the null point is found at a distance of l1 cm from A. If a resistance of X is connected in parallel with S, then null point occurs at a distance l2 cm from A Obtain the formula for X in terms of l1,l2 and S.

Ans.

Ans.

Ans.(i)

27.(i) State the principle of working of a meter bridge.
(ii) In a meter bridge balance point is found at a distance l1 with resistances R and S as shown in the figure. When an unknown resistance X is connected in parallel with the resistance S, the balance point shifts to a distance l₂. Find expression for X in terms of l1 , l₂ and S. [All India 2009]

Ans.

Ans.

29.State Kirchhoff’s rules. Use Kirchhoff’s rules to show that no current flows in the given circuit

Ans.

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31.Draw a circuit showing a Wheatstone bridge. Use Kirchhoff’s rule to obtain the balance condition in terms of the values of the four resistors for the galvanometer to give null deflection.  [Delhi 2008 CI
Ans.

32.Draw a circuit diagram for a Wheatstone bridge. Explain briefly how the balance condition for the galvanometer to give null deflection provides a practical method for the determination of an unknown resistance? [Delhi 2008]
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5 Marks Questions

33.(i) State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge.
(ii) In the meter bridge experimental set up, shown in the figure, the null point D is obtained at a distance of 40 cm from end A of the meter bridge wire.

Ans.

34.(i) Use Kirchhoff’s rules to obtain the balance condition in a Wheatstone bridge.
(ii) Calculate the value of R in the balance condition of the  Wheatstone bridge, if the carbon resistor connected across the arm CD has the colour sequence red, red and orange, as shown in the figure.

(iii) If now the resistance of the arms BC and CD are interchanged, to obtain the balance condition, another carbon resistor is connected in place or R. What would now be sequence of colour bands of the carbon resistor?[Delhi 2012]
Ans.

35.(i) State with the help of a circuit diagram, the working principle of a meter bridge. Obtain the expression used for determining the unknown resistance.
(ii)What happens if the galvanometer and cell are interchanged at the balance point of the bridge?
(iii)Why is it considered important to obtain the balance point near the mid-point of the wire? [Delhi 2011 c]
Ans.(i)

(ii)

(iii)It is because of the fact that meter bridge is most sensitive when null point occur nc the mid-point of wire and all the four resistances are of same order.

36.State the two rules that serves as general rules for analysis of electrical circuits.Use these rules to write the three equations that may be used to obtain the values of the three unknown currents in the branches (shown) of the circuit given below. [All India 2008 C]

Ans.

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