Question

A conductor carrying 14.7 amps of current is directed along the positive x-axis and perpendicular to a uniform magnetic field. A magnetic force per unit length of 0.125 n/m acts on the conductor in the negative y direction. What is the strength of the magnetic field at the place where the current is?

Answer 1

Answer:

0.0085 T

Explanation:

The magnetic force per unit length exerted on a current-carrying wire is given by

$\frac{F}{L}=BI sin \theta$

where

B is the strength of the magnetic field

I is the current

$\theta$ is the angle between the directions of B and I

In this problem, we know

$\frac{F}{L}=0.125 N/m$

I = 14.7 A (the current)

$\theta=90^{\circ}$ since I and B are perpendicular

Therefore, solving for B, we find the strength of the field:

$B=\frac{F/L}{I sin \theta}=\frac{0.125}{(14.7)(1)}=0.0085 T$