Question:

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 :

Show Hint

Motional emf = B l v with B = mu0 I / (2 pi r) at the connector; divide by R (the only resistance).
Updated On: Nov 5, 2026
  • \(\frac{\mu_0}{4\pi}\frac{Ivl}{Rr}\)
  • \(\frac{\mu_0}{\pi}\frac{Ivl}{Rr}\)
  • \(\frac{\mu_0}{2\pi}\frac{Ivl}{Rr}\)
  • \(\frac{2\mu_0}{\pi}\frac{Ivl}{Rr}\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Magnetic field of the long wire at distance r: \(B = \frac{\mu_0 I}{2\pi r}\).
The sliding connector of length \(l\) moves with speed v perpendicular to B, so the motional emf is \(\varepsilon = B l v = \frac{\mu_0 I l v}{2\pi r}\).
The only resistance in the loop is that of the connector (R), so the induced current is
\(i = \frac{\varepsilon}{R} = \frac{\mu_0}{2\pi}\frac{I v l}{R r}\).
Answer: (3) \(\frac{\mu_0}{2\pi}\frac{Ivl}{Rr}\)
Was this answer helpful?
0
0

Top JEE Main Physics Questions

View More Questions

Top JEE Main Electromagnetic induction Questions

View More Questions