A) 14.85 W
B) 29.7 W
C) 1.57 W
D) 2.97 W
Correct Answer: C
Solution :
Force on conductor = I L B \[B={{B}_{0}}{{e}^{-\alpha x}}\] (Due to magnetic field) [+ x−direction ] \[\left\langle P \right\rangle =\frac{1}{T}\int\limits_{0}^{T}{F.V.}dt=\frac{1}{T}\int\limits_{0}^{{{x}_{0}}}{F}.dx\]\[[{{x}_{0}}=2m]\] \[\Rightarrow \]\[\left\langle P \right\rangle =\left[ \frac{1}{T}\frac{I\,L\,{{B}_{0}}}{(-\alpha )}{{e}^{-\alpha x}} \right]_{0}^{2}\] \[\Rightarrow \]\[\left\langle P \right\rangle =\frac{I\,L\,{{B}_{0}}}{T\times 0.2}\left[ 1-{{e}^{-\alpha (2m)}} \right]\] \[=\frac{10\times 3\times 3\times {{10}^{-6}}}{5\times {{10}^{-3}}\times 0.2}\left[ 1-{{e}^{-0.4}} \right]\] \[=\frac{1.8}{0.2}\left( 1-\left( 1-(0.4)+\frac{0.16}{2}-\frac{0.064}{16}+...... \right) \right)\] \[\approx 9\times \frac{0.64}{2}=2.88\] On Exact evaluation \[\left\langle P \right\rangle =2.97W\]You need to login to perform this action.
You will be redirected in
3 sec