Question

A rectangular loop has a sliding connector $PQ$ 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 $I_{1},I_{2} \, and \, I$ are

• A. $I_{1}=-I_{2}=\frac{B l v}{R}, \, I=\frac{2 B l v}{R}$
• B. $I_{1}=I_{2}=\frac{B l v}{3 R},I=\frac{2 B l v}{3 R}$
• C. $I_{1}=I_{2}=I=\frac{B l v}{R}$
• D. $I_{1}=I_{2}=\frac{B l v}{6 R},I=\frac{B l v}{3 R}$

Answer 1

Answer: (B)

A moving conductor is equivalent to battery of emf
$ \, \, =vBl$ (motion emf)
Equivalent circuit

$ \, \, I=I_{2}+I_{2}$
Applying Kirchhoff's law
$I_{1}R+IR-vBl=0 \, \, \ldots \left(\right.i\left.\right)$
$I_{2}R+IR-vBl=0$ ...(ii)
Adding Eqs. (i) and (ii), we get
$ \, \, 2IR+IR=2vBl$
$ \, \, \, I=\frac{2 v B l}{3 R}$
$ \, I_{1}=I_{2}=\frac{v B l}{3 R}$