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Q.

A small coil C with N= 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in figure. The area of the coil is S = 1cm2, the length of the right arm of the balance beam is I = 30 cm. When there is no current in the coil the balance is in equilibrium. On passing a current I = 22 mA through the coil, equilibrium is restored by putting an additional weight of mass m = 60 mg on the balance pan. Find the magnetic induction field (in terms of x 10-1 T) between the poles of the electromagnet, assuming it to be uniform.

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answer is 4.

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Detailed Solution

Magnetic torque =NISBsin(S→,B→)  NISB sin90°  Gravitational torque =(Δm×g)l  For equilibrium NISB=Δmgl⇒ B=Δmgl/NIS B=60×10-6×9.8×0.3200×22×10-3×1×10-4=4×10-1 T
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A small coil C with N= 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in figure. The area of the coil is S = 1cm2, the length of the right arm of the balance beam is I = 30 cm. When there is no current in the coil the balance is in equilibrium. On passing a current I = 22 mA through the coil, equilibrium is restored by putting an additional weight of mass m = 60 mg on the balance pan. Find the magnetic induction field (in terms of x 10-1 T) between the poles of the electromagnet, assuming it to be uniform.