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 I1, I2 and I are

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 I1, I2 and I are (A) I1 = -I2 = (Bℓ v/R), I = (2Bℓ v/R) (B) I1 = -I2 = (Bℓ v/3R), I = (2Bℓ v/3R) (C) I1 = -I2 = I = (Bℓ v/R) (D) I1 = -I2 = (Bℓ v/6R), I = (Bℓ v/3R)

Tardigrade Admin · Accepted Answer

A moving conductor is equivalent to a battery of

e m f = v B ℓ

(motion emf)
Equivalent circuit

I = I_{1} + I_{2}

applying Kirchoff’s law

I_{1} R + I R - v B ℓ = 0

______(1)

I_{2} R + I R - v B ℓ = 0

______(2)
adding

(1)

&

(2)

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}

Q. 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

$I_{1} = - I_{2} = \frac{B ℓ v}{R}, I = \frac{2 B ℓ v}{R}$

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

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

$I_{1} = - I_{2} = \frac{B ℓ v}{6 R}, I = \frac{B ℓ v}{3 R}$

Q. A rectangular loop has a sliding connector PQ of length ℓ and resistance RΩ 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 I1​,I2​ and I are

I1​=−I2​=RBℓv​,I=R2Bℓv​

I1​=−I2​=3RBℓv​,I=3R2Bℓv​

I1​=−I2​=I=RBℓv​

I1​=−I2​=6RBℓv​,I=3RBℓv​

A moving conductor is equivalent to a battery of emf=vBℓ (motion emf) Equivalent circuit I=I1​+I2​ applying Kirchoff’s law I1​R+IR−vBℓ=0 ______(1) I2​R+IR−vBℓ=0 ______(2) adding (1) & (2) 2IR+IR=2vBℓ I=3R2vBℓ​ I1​=I2​=3RvBℓ​

Q. 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

$I_{1} = - I_{2} = \frac{B ℓ v}{R}, I = \frac{2 B ℓ v}{R}$

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

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

$I_{1} = - I_{2} = \frac{B ℓ v}{6 R}, I = \frac{B ℓ v}{3 R}$