JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Electromagnetic Induction

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An infinitely long straight wire carrying current I, one side opened rectangular loop and a conductor C with a sliding connector are located in the same plane, as shown in the figure. The connector has length l and resistance R. It slides to the right with a velocity v . The resistance of the conductor and the self inductance of the loop are negligible. The induced current in the loop, as a function of separation r , between the connector and the straight wire is:

(A) $\frac{μ _{0}}{2 π} \frac{Ivl}{Rr}$ ✓ Correct
(B) $\frac{μ _{0}}{π} \frac{Iv l}{Rr}$
(C) $\frac{2 μ _{0}}{2 π} \frac{Ivl}{Rr}$
(D) $\frac{μ _{0}}{4 π} \frac{Ivl}{Rr}$

Answer: (A)

Solution

The long straight wire sets up a field

B = \frac{μ _{0} I}{2 π r}

at distance

r

. The simplest route to the emf is the motional picture: the sliding rod of length

ℓ

moves at speed

v

through the local field, so

ε = B ℓ v = \frac{μ _{0} I ℓ v}{2 π r}

. Dividing by the rod's resistance

R

, the induced current when the connector is a distance

r

from the wire is

i = \frac{ε}{R} = \frac{μ _{0}}{2 π} \frac{I v ℓ}{R r}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Electromagnetic Induction

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