First slide

Torque on a current carrying loop placed in uniform magnetic field

Question

A uniform current carrying ring of mass m and radius R is connected by a massless string as shown in Fig. A uniform magnetic field $B_{0}$ exists in the region to keep the ring in horizontal position, then the current in the ring is (l = length of string)

Easy

A

$\frac{m g}{π R B_{0}}$
B

$\frac{m g}{R B_{0}}$
C

$\frac{m g}{3 π R B_{0}}$
D

$\frac{m g l}{π R^{2} B_{0}}$

Solution

Torque due to magnetic field = $i B_{0} . π R^{2}$
Torque due to the mass = $m g . R$
Now $m g . R = i B_{0} . π R^{2} \Rightarrow i = \frac{m g}{π B_{0} R}$

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