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  • A rectangular loop has a sliding connector PQ of length l and resistnace  and i

A rectangular loop has a sliding connector PQ of length l and resistnace \mathrm{R\Omega } and it is moving with a speed v as shown. The set up is placed in a uniform magnetic field going into the plane of the paper. The three currents \mathrm{I_{1},I_{2}} and I are:

Option: 1

\mathrm{I_1=I_2=\frac{B l v}{6 R}, I=\frac{B l v}{3 R}}


Option: 2

\mathrm{I}_1=-\mathrm{I}_2=\frac{\mathrm{B} l \mathrm{v}}{\mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} l \mathrm{v}}{\mathrm{R}}


Option: 3

\mathrm{I_1=I_2=\frac{B l v}{3 R}, I=\frac{2 B l v}{3 R}}


Option: 4

\mathrm{I}_1=\mathrm{I}_2=\mathrm{I}=\frac{\mathrm{B} / \mathrm{V}}{\mathrm{R}}


Answers (1)

best_answer

Circuit can be reduced as

\mathrm{I}=\frac{\mathrm{e}}{3 \mathrm{R} / 2}=\frac{2 \mathrm{v} / \mathrm{B}}{3 \mathrm{R}} \Rightarrow \mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{I}}{2}=\frac{\mathrm{v} / \mathrm{B}}{3 \mathrm{R}}

 

Posted by

Suraj Bhandari

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