A rectangular loop has a sliding connector $P Q$ of length $l$ and resistance $R Ω$ 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 $I_{1}, I_{2} a n d I$ are

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TestHub · Accepted Answer

A moving conductor is equivalent to battery of emf

$= v B l$ (motion emf)

Equivalent circuit

$I = I_{2} + I_{2}$

Applying Kirchhoff's law

$I_{1} R + I R - v B l = 0 \dots (i)$

$I_{2} R + I R - v B l = 0$ ...(ii)

Adding Eqs. (i) and (ii), we get

$2 I R + I R = 2 v B l$

$I = \frac{2 v B l}{3 R}$

$I_{1} = I_{2} = \frac{v B l}{3 R}$

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