Exams › JEE Advanced › Physics

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?

1. μ₀l ln(b/a) / 2π
2. μ₀l ln(a/b) / π
3. μ₀l ln(a/b) / 2π
4. μ₀l ln(b/a) / π

Correct answer: μ₀l ln(b/a) / 2π

Solution

The self-inductance of a coaxial cable depends on the logarithmic ratio of the radii of the conductors and the length of the cable. The correct expression is μ₀l ln(b/a) / 2π.

Related JEE Advanced Physics questions

• What is the final velocity attained by the loop?
• 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.
• A uniform magnetic field B = (1/pi)*sin(2t) tesla directed into the plane of the paper exists in a cylindrical region of radius 1 m. An equilateral triangular loop of total resistance 10 ohm and side length 2 m is placed symmetrically around the cylindrical field region. Find the maximum potential difference across side AB of the triangle.

⚔️ Practice JEE Advanced Physics free + battle 1v1 →