Question

An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic field on the surface of the wire is measured to be B1. The experiment is then repeated with the same current but with a wire of diameter 2d. The magnetic field measured on the surface of this second wire will be which of the following?

B1, B1 / 4, 4B1, B1 / 2

Answer 1

Answer:

$B_2 = \frac{B_1}{2}$

Explanation:

As per Ampere's law the magnetic field at the surface of the wire is given as

$\int B. dL = \mu_0 i$

here we have

$B . (\pi d) = \mu_0 i$

so we will have

$B_1 = \frac{\mu_0 i}{\pi d}$

now again we use same value of current but wire with double the diameter

so the magnetic field at the surface is given as

$B_2 = \frac{\mu_0 i}{2\pi d}$

so we have

$B_2 = \frac{B_1}{2}$