JEE Main · 2010 · 11 Apr 2010 · Offline · Electromagnetic Induction

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

Answer: (B)

Solution

The sliding rod PQ is the only part cutting field lines, so it acts as a cell of motional emf

ε = B ℓ v

with its own resistance

R

. The two arms of the rectangle on either side present two resistances

R

in parallel across the rod, an external resistance of

R /2

. The current the rod drives is

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

, and by symmetry it splits equally between the two identical arms, so

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

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option