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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 :

23871_5e0e475ebf4346a8be7c5b98a62aedfc.png


A
40π rad/s2
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B
20π rad/s2
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C
5π rad/s2
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D
15π rad/s2
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Solution

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

Physics
NCERT
Standard XII

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