A rectangular conducting loop of length 4 cm and width 2 cm is in the xy-plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction 23x^+21y^ with a constant speed v. The wire is carrying a steady current I=10A in the positive x-direction. A current of 10μA flows through the loop when it is at a distance d=4cm from the wire. If the resistance of the loop is 0.1Ω, then the value of v is ____ ms−1. [Given: The permeability of free space μ0=4π×10−7NA−2 ]
Answer: 4
Solution
The magnetic field from a long wire at distance r is B=2πrμ0I.
The induced emf is due to the change in flux as the loop moves. The component of velocity perpendicular to the wire is vy=vsin(30°)=v/2.
The emf induced is: ε=(B1−B2)bvy=2πμ0I(d1−d+a1)bvy
where a=4 cm (length), b=2 cm (width), d=4 cm.
Current: i=Rε
10×10−6=0.12×10−7×10(41−81)×2×vy×10−2
Solving: vy=2 m/s.
Since vy=v/2, we get v=4 m/s.
JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer