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A conducting ring of mass 2 Kg and radius $0.5m$ is placed on a smooth horizontal surface. The ring carries a current $i = 10A$. A horizontal magnetic field $B = 4T$ is now switched on. The angular acceleration of the ring will be

$\begin {array} {1 1} (a)\;10 \pi \: rad/s^2 & \quad (b)\;20 \pi \: rad/s^2 \\ (c)\;30 \pi \: rad/s^2 & \quad (d)\;40 \pi \: rad/s^2 \end {array}$

 

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$ \overrightarrow M = i\: A\: \hat K = 10 \pi (0.5)^2 \hat K$
$ \tau= \overrightarrow M \times \overrightarrow B = 2.5 \pi \times 4 = 10 \pi$
Now $\tau=I \alpha $
 $I= M R^2  =2 (0.5)^2 =0.5$
hence angular acceleration $\alpha = \tau / I = 20 \pi$
 
Ans : (b)

 

answered Feb 19, 2014 by thanvigandhi_1
edited Sep 15, 2014 by thagee.vedartham
 

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