A metal rod of resistance R is fixed along a diameter of a conducting ring of radius r. There is a magnetic field of magnitude B perpendicular to the plane of the loop. The ring spins with angular velocity ω about its axis. The centre of the ring is joined to its rim by external wire W. The ring and W have no resistance. The current in W is

Question

A metal rod of resistance R is fixed along a diameter of a conducting ring of radius r. There is a magnetic field of magnitude B perpendicular to the plane of the loop. The ring spins with angular velocity ω about its axis. The centre of the ring is joined to its rim by external wire W. The ring and W have no resistance. The current in W is (A) (B r2 ω/R) (B) (B r2 ω/2 R) (C) (2 B r2 ω/R) (D) zero

Tardigrade Admin · Accepted Answer

ε=(B r2 ⋅ ω/2) Each half of metal rod has a resistance (R/2) . Both halves are in parallel, so Re q=(R/4) ⇒ I=(ε/Re q)=(B r2 ω/2((R)4))=(2 B r2 ω/R)

$\frac{B r ^{2} ω}{R}$

$\frac{B r ^{2} ω}{2 R}$

$\frac{2 B r ^{2} ω}{R}$

zero

$ε = \frac{B r ^{2} \cdot ω}{2}$
Each half of metal rod has a resistance $\frac{R}{2} .$ Both halves are in parallel, so
$R_{e q} = \frac{R}{4}$
$\Rightarrow I = \frac{ε}{R _{e q}} = \frac{B r ^{2} ω}{2 ( \frac{R}{4} )} = \frac{2 B r ^{2} ω}{R}$

RBr2ω​

2RBr2ω​

R2Br2ω​

zero

ε=2Br2⋅ω​ Each half of metal rod has a resistance 2R​. Both halves are in parallel, so Req​=4R​ ⇒I=Req​ε​=2(4R​)Br2ω​=R2Br2ω​

$\frac{B r ^{2} ω}{R}$

$\frac{B r ^{2} ω}{2 R}$

$\frac{2 B r ^{2} ω}{R}$

$ε = \frac{B r ^{2} \cdot ω}{2}$
Each half of metal rod has a resistance $\frac{R}{2} .$ Both halves are in parallel, so
$R_{e q} = \frac{R}{4}$
$\Rightarrow I = \frac{ε}{R _{e q}} = \frac{B r ^{2} ω}{2 ( \frac{R}{4} )} = \frac{2 B r ^{2} ω}{R}$