A) \[\frac{Bi{{\ell }^{2}}}{x}\]
B) \[\frac{Bi{{\ell }^{2}}}{4x}\]
C) \[\frac{Bi{{x}^{2}}}{\ell }\]
D) \[\frac{Bi{{x}^{2}}}{4\ell }\]
Correct Answer: B
Solution :
Magnetic force on the wire |
\[F=Bi\frac{\ell }{2}\sin \theta \simeq Bi\frac{\ell }{2}\tan \theta \] |
\[=Bi\frac{\ell }{2}\times \frac{\ell /2}{x}\] |
\[=\frac{Bi{{\ell }^{2}}}{4x}\] |
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