Exams › JEE Advanced › Physics

A rectangular loop contains a sliding conducting rod of length l = 1.0 m in a uniform magnetic field B = 2 T perpendicular to the plane of the loop. The rod's resistance is r = 2 ohm, and two resistors of 6 ohm and 3 ohm are connected (in parallel with each other across the rod). Find the external force needed to move the rod at constant velocity v = 2 m/s.

1. 2 N
2. 0.5 N
3. 4 N
4. 8 N

Correct answer: 2 N

Solution

EMF = Blv = 2*1*2 = 4 V. The 6 ohm and 3 ohm in parallel give 2 ohm, in series with r = 2 ohm, so R_total = 4 ohm. Current through the rod I = 4/4 = 1 A. For constant velocity the external force balances the magnetic force: F = B*I*l = 2*1*1 = 2 N.

Related JEE Advanced Physics questions

• What is the final velocity attained by the loop?
• A long coaxial cable is made up of two cylindrical conductors with radii a and b. The inner conductor carries a constant current i, while the outer conductor acts as the return path for the current. What is the self-inductance of a segment of the cable of length l?
• Determine the inductance per unit length for a loop created by connecting the ends of two infinitely long parallel wires, each with radius r, and separated by a distance d between their centers. Ignore the magnetic flux inside the wires and the effects at the ends.
• If the electromotive force is 10 V and the rate of change of current is calculated as 2.5 A/s, what is the self-inductance of the coil, given by L = e / (dl/dt)?
• The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is
• A semicircular loop of radius r is positioned on the axis of a large current-carrying circular ring of radius R (with R >> r). The loop is placed at an axial distance of R*sqrt(3) from the center of the ring, and the plane of the semicircular loop makes an angle of 53 deg with the plane of the ring. Find the mutual inductance M between the two coils.

⚔️ Practice JEE Advanced Physics free + battle 1v1 →