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  • A rod of length b moves with a constant velocity v in the magnetic field of a straight long conductor that carries a current I as shown in the figure. The emf induced in the rod is <img a

A rod of length b moves with a constant velocity v in the magnetic field of a straight long conductor that carries a current I as shown in the figure. The emf induced in the rod is

Option: 1

\mathrm{\frac{\mu_0 i v}{2 \pi} \tan ^{-1} \frac{a}{b}}


Option: 2

\mathrm{\frac{\mu_0 i v}{2 \pi} \ln \left(1+\frac{b}{a}\right)}


Option: 3

\mathrm{\frac{\mu_0 i v \sqrt{a b}}{4 \pi(a+b)}}


Option: 4

\mathrm{\frac{\mu_0 i v(a+b)}{4 \pi a b}}


Answers (1)

best_answer

The induced emf between two ends of a segment dx = dE = Bvdx

 Where B = magnetic field due to straight current carrying wire at the segment \mathrm{d x=\frac{\mu_0 i}{2 \pi x}}

\mathrm{\Rightarrow d E=\frac{\mu_0 i v d x}{2 \pi x}}

The induced emf between the ends of the rod

\mathrm{\begin{aligned} & =E=\int d E=\frac{\mu_0 i v}{2 \pi} \int_a^{a+b} \frac{d x}{x} \\ & \Rightarrow E= \frac{\mu_0 i \mathrm{v}}{2 \pi} \ln \left(\frac{b}{a}+1\right) \\ \end{aligned}}

Posted by

HARSH KANKARIA

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