Question

A long straight cylindrical shell carries current i parallel to its axis and uniformly distributed over its cross section. The magnitude of the magnetic field is greatest:

A) at the inner surface of the shell
B) at the outer surface of the shell
C) inside the shell near the middle
D) in hollow region near the inner surface of the shell
E) near the center of the hollow region

Answer 1

The magnetic field is zero within the hollow cavity of a cylindrical shell carrying current. It is greatest at the inner surface of the shell when the distance 'r' equals the inner radius 'r1' of the shell.

Step-by-step explanation:

The question asks about the magnetic field around a current-carrying cylindrical shell. According to Ampère's Law and the Biot-Savart Law, the magnetic field B inside the cavity of a long straight cylindrical shell is zero. This is because the currents that produce the magnetic field through each 'ring' of the cylindrical shell cancel each other out within the cavity. Therefore, the magnetic field is not greatest in the hollow region. Outside the shell, the magnetic field is created by the entire current and the magnitude increases with closer proximity to the current source.

For a long, hollow, cylindrical conductor with a uniformly distributed current, the magnetic field within the hollow region is zero and at any radial distance r from the center of the conductor that is less than the inner radius r1. Therefore, the magnetic field's magnitude is greatest just at the inner surface of the shell, where r equals r1.