Class 12 Physics

CBSE and UP Board

Moving Charges and Magnetism

Chapter-4 Exercise 4.24

Additional Exercise

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Magnetic field strength, B = 3000 G = 3000 x 10

^{-4}T = 0.3 TLength of the rectangular loop,

l=10 cmWidth of the rectangular loop, b = 5 cm

Area of the loop, A =

lx b = 10×5 = 50 cm^{2}= 50 x 10⁻⁴m²Current 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⁻

^{2}j NmThe torque is 1.8×10

^{2}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 NmThe 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⁻

^{4}x0.3 = 1.8×10⁻^{2}NmThe 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⁻

^{4}x 12 ) k x 0.3 k = 0Hence, 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.