Exams › JEE Advanced › Physics

A magnetic field points into the plane of the page. A rod of length l moves in this field so that its bottom end has velocity v1 and its top end has velocity v2 (with v2 > v1). The emf induced across the rod is:

1. B*v1*l
2. B*v2*l
3. (1/2)*B*(v2 + v1)*l
4. (1/2)*B*(v2 - v1)*l

Correct answer: (1/2)*B*(v2 + v1)*l

Solution

Since the rod's velocity varies linearly from v1 (bottom) to v2 (top), the average velocity is (v1 + v2)/2. Motional emf = B * (average velocity) * l = (1/2)*B*(v1 + v2)*l.

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 →