Question

A wire carries a current of 10 amps. a researcher finds that the strength of the induced magnetic field at a certain point is 0.50 g. what will the strength of the induced field be at the same point if the current is increased to 25 amps?

Answer 1

The strength of the magnetic field produced by a current-carrying wire at a distance r from the wire is
$B(r)= \frac{\mu_0 I}{2 \pi r}$
where I is the current in the wire.

From the equation, we see that the strength of the field is directly proportional to the current intensity, I. In the problem, the current is increased from 10 A to 25 A, so by a factor
$\frac{I_2}{I_1}= \frac{25 A}{10 A}=2.5$
therefore, the strength of the magnetic field will increase by the same factor:
$B_2 = 2.5 B_1 = 2.5 (0.50 G)=1.25 G$
And the strength of the new magnetic field is 1.25 G.