A) \[1.5\times {{10}^{-6}}\,A\] anticlockwise
B) \[1.5\times {{10}^{-6}}\,A\]clockwise
C) \[0.75\times {{10}^{-6}}\,A\] anticlockwise
D) \[0.75\times {{10}^{-6}}\,A\]clockwise
Correct Answer: A
Solution :
[a] Rate of change of area of the loop \[\frac{dA}{dt}=A\], \[\beta \frac{dT}{dt}=A.(2\alpha )\frac{dT}{dt}=\frac{3}{4}\times 2\times {{10}^{-6}}\times 1\] \[=11.5\times {{10}^{-6}}{{m}^{2}}/s\] \[emf=-\frac{d\phi }{dt}=-\frac{\beta .dA}{dt}=-1.5\times {{10}^{-6}}V\] current in the loop\[=1.5\times {{10}^{-6}}A\] The direction will be anticlockwise as the induced current will try to negate the increase in fluix due to increase in area.You need to login to perform this action.
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