A) zero
B) \[{{v}_{0}}{{B}_{0}}d\]
C) \[\frac{{{v}_{0}}{{B}_{0}}{{d}^{3}}}{{{a}^{2}}}\]
D) \[\frac{{{v}_{0}}{{B}_{0}}{{d}^{2}}}{a}\]
Correct Answer: D
Solution :
[d] At \[x=0\], \[B={{B}_{0}}\], so \[{{e}_{1}}={{B}_{0}}{{v}_{0}}d\] At,\[x=d\],\[B={{B}_{0}}\left( 1+\frac{d}{a} \right)\], so \[{{e}_{2}}={{B}_{0}}\left( 1+\frac{d}{a} \right){{v}_{0}}d\] Now \[{{e}_{net}}={{e}_{2}}-{{e}_{1}}=\frac{{{B}_{0}}{{v}_{0}}{{d}^{2}}}{a}\]You need to login to perform this action.
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