Question

A coil of wire carrying a current may be used to create a magnetic field at the center of the coil. If the radius of the coil is doubled while the number of windings and the current remain the same, then how does this change the strength of the magnetic field at the center?

Answer 1

Answer:

Here magnetic field at center does not change,  it remains constant .

Explanation:

The magnetic field at the center of the solenoid depends on only the current flowing through the wire and the number of turns of the coil around the solenoid .

  Since magnetic field at the center of the solenoid is given by the expression

                           $B=\mu_{o}ni$

  Here  $\mu_{o}$ is permeability constant whose value is equal to $4\pi\times 10^{-7}\frac{H}{m}$  .

 N is number of turns of the coil .

  I is current flowing through the coil .

The field lines near the center of the solenoid are approximately parallel , indicating a nearly uniform magnetic field.