A) \[50\mu A\]
B) \[114\mu A\]
C) \[{{l}_{1}}\]
D) \[{{T}_{1}}\]
Correct Answer: A
Solution :
Given, n = 500 \[\left( \because a=\frac{dv}{dt} \right)\] and \[F'=ilB=C{{B}^{2}}{{I}^{2}}a\] The magnetic flux linked with the coil is \[F-F'=ma\Rightarrow F=ma+C{{B}^{2}}{{I}^{2}}a\] \[a=\frac{F}{m+C{{B}^{2}}{{I}^{2}}}\].You need to login to perform this action.
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