Question

Inside a long straight uniform wire of round cross-section, there is a long round cylindrical cavity whose axis is parallel to the axis of the wire and displaced from latter by a distance d. If a direct current of density ¯J flows along the wire, then magnetic field inside the cavity will be:

A 0
B 12μ0¯Jׯd
C μ0¯Jׯd
D 32μ0¯Jׯd

Answer 1

The magnetic field inside the cylindrical cavity of a wire is given by B = μ0 J × d, where J is the current density and d is the displacement between the cavity and the wire.

Therefore, the correct option is C, μ0J × d.

Step-by-step explanation:

The magnetic field inside the cylindrical cavity can be determined by superimposing the field of the wire onto the field of the cavity. The magnetic field inside the cavity is given by the formula B = μ0 J × d, where μ0 is the permeability of free space, J is the current density, and d is the displacement between the cavity and the wire.

Since the current density in the wire is given as ¯J and it is along the axis of the wire, the magnetic field inside the cavity will be in the same direction as ¯J × d. Therefore, the correct option is C, μ0J × d.