Question

# A conducting ring of mass 2 kg and radius 0.5m is placed on a smooth horizontal plane. The ring carries a current $$i = 4A$$. A horizontal magnetic field $$B = 10T$$ is switched on at $$t = 0$$ as shown in figure. The initial angular acceleration is :

A
B
C
D

Solution

## The correct option is B $$20\pi \ rad/s^{2}$$Torque on the ring is,$$\vec{\tau} =\vec{M}\times\vec{B}$$$$\vec {\tau}=MBSin\theta$$       $$=i\pi r^2B Sin 90$$       $$=4\times\pi\times0.25\times10\times1$$       $$=10\pi\ Nm$$$$\vec{\tau}=I\alpha$$$$10\pi=Mr^2\alpha$$$$\alpha=\dfrac{10\pi}{2\times0.25}$$$$=20\pi \ rad/s^2$$PhysicsNCERTStandard XII

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