A rectangular loop with a sliding conductor of length l is located in a uniform magnetic field perpendicular to the plane of the loop (figure). The magnetic induction is B. The conductor has a resistance R. The sides A B and C D have resistances R1 and R2, respectively. Find the current through the conductor during its motion to the right with a constant velocity v.

Question

A rectangular loop with a sliding conductor of length l is located in a uniform magnetic field perpendicular to the plane of the loop (figure). The magnetic induction is B. The conductor has a resistance R. The sides A B and C D have resistances R1 and R2, respectively. Find the current through the conductor during its motion to the right with a constant velocity v. (A) (B l v(R1+R2)/R1(R1+R2)) (B) (B l2 v/R1+R1 R2) (C) (B l v(R1+R2)/R1 R2+R(R1+R2)) (D) (B l2 v/R1 R2+R(R1+R2))

Tardigrade Admin · Accepted Answer

Induced emf =B l v . R is internal resistance of seat of emf, i.e., of rod Total resistance of circuit =R+(R1 R2/R1+R2) ∴ I=(B l v/R+(R1 R2)(R1+R2)) =(B l v(R1+R2)/R1 R2+R(R1+R2))

$\frac{Bl v ( R _{1} + R _{2} )}{R _{1} ( R _{1} + R _{2} )}$

$\frac{B l ^{2} v}{R _{1} + R _{1} R _{2}}$

$\frac{Bl v ( R _{1} + R _{2} )}{R _{1} R _{2} + R ( R _{1} + R _{2} )}$

$\frac{B l ^{2} v}{R _{1} R _{2} + R ( R _{1} + R _{2} )}$

Induced emf $= Bl v . R$ is internal resistance of seat of emf, i.e., of rod
Total resistance of circuit $= R + \frac{R _{1} R _{2}}{R _{1} + R _{2}}$
$∴ I = \frac{Bl v}{R + \frac{R _{1} R _{2}}{( R _{1} + R _{2} )}}$
$= \frac{Bl v ( R _{1} + R _{2} )}{R _{1} R _{2} + R ( R _{1} + R _{2} )}$

R1​(R1​+R2​)Blv(R1​+R2​)​

R1​+R1​R2​Bl2v​

R1​R2​+R(R1​+R2​)Blv(R1​+R2​)​

R1​R2​+R(R1​+R2​)Bl2v​

Induced emf =Blv.R is internal resistance of seat of emf, i.e., of rod Total resistance of circuit =R+R1​+R2​R1​R2​​ ∴I=R+(R1​+R2​)R1​R2​​Blv​ =R1​R2​+R(R1​+R2​)Blv(R1​+R2​)​

$\frac{Bl v ( R _{1} + R _{2} )}{R _{1} ( R _{1} + R _{2} )}$

$\frac{B l ^{2} v}{R _{1} + R _{1} R _{2}}$

$\frac{Bl v ( R _{1} + R _{2} )}{R _{1} R _{2} + R ( R _{1} + R _{2} )}$

$\frac{B l ^{2} v}{R _{1} R _{2} + R ( R _{1} + R _{2} )}$

Induced emf $= Bl v . R$ is internal resistance of seat of emf, i.e., of rod
Total resistance of circuit $= R + \frac{R _{1} R _{2}}{R _{1} + R _{2}}$
$∴ I = \frac{Bl v}{R + \frac{R _{1} R _{2}}{( R _{1} + R _{2} )}}$
$= \frac{Bl v ( R _{1} + R _{2} )}{R _{1} R _{2} + R ( R _{1} + R _{2} )}$