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Home/ Questions/Q 7710
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Ashok0210
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Ashok0210
Asked: February 23, 20212021-02-23T12:02:58+00:00 2021-02-23T12:02:58+00:00In: Class-11-Physics

A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

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A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

Class 12 Physics
CBSE and UP Board
Moving Charges and Magnetism
Chapter-4 Exercise 4.24
Additional Exercise

2020-2021cbse and up boardclass 12 chapter-4 physicsclass 12 physicsncert class 12 moving charges and magnetism
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    1. Pawan1308
      2021-02-27T13:06:04+00:00Added an answer on February 27, 2021 at 1:06 pm

      Magnetic field strength, B = 3000 G = 3000 x 10-4 T = 0.3 T

      Length of the rectangular loop, l=10 cm

      Width of the rectangular loop, b = 5 cm

      Area of the loop, A = l x b = 10×5 = 50 cm2 = 50 x 10⁻⁴m²

      Current in the loop, I = 12 A

      Now, taking the anti-clockwise direction of the current as positive and vice-versa:

      Ans (a).

      Torque, τ= IA x B   (τ,A & B in vector forms)

      From the given figure, it can be observed that A is normal to the y-z plane and B is directed along the z-axis.

      Therefore,

      τ = 12 x (50 x 10⁻⁴ ) i x 0.3k  (i ,j & k in vector forms)

      = -1.8×10⁻2j Nm

      The torque is 1.8×102 N m along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.

      Ans (b).

      This case is similar to case (a). Hence, the answer is the same as (a).

      Ans (c).

      Torque,  τ= IAxB            (τ,A & B in vector forms)

      From the given figure, it can be observed that A is normal to the x-z plane and B is directed along the z-axis.

      Therefore ,

      τ= -12 x (50 x 10⁻⁴ ) i x 0.3k  (i ,j & k are vectors)

      = —1.8×10-2 i Nm

      The torque is 1.8×10-2 Nm along the negative x direction and the force is zero.

      Ans (d).

      Magnitude of torque is given as:

      |τ| = IAB

      = 12x50x10⁻4x0.3 = 1.8×10⁻2Nm

      The torque is 1.8×10⁻2 N m at an angle of 240° with positive x direction. The force is zero.

      Ans (e).

      Torque, τ = IAxB = (50×10⁻4 x 12) k x 0.3k = 0          (τ,A & B in vector forms) and (i ,j & k are vectors)

      Hence, the torque is zero. The force is also zero.

      Ans (f).

      Torque τ = IAxB            (τ,A & B in vector forms) and (i ,j & k are vectors)

      = (50×10⁻4x 12 ) k x 0.3 k = 0

      Hence, the torque is zero. The force is also zero.

      In case (e), the direction of IA and B is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.

      Whereas, in case (f), the direction of IA and B is opposite. The angle between them is 180°. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.

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