Question

A horizontal magnetic field of 0.075 T is at an angle of 19.015o to the direction of the current in a straight, horizontal wire 64.714 cm long. If the wire carries a current of 8.321 A, what is the magnitude of the force on the wire ?

Answer 1

Given:

The magnetic field is B = 0.075 T

The length of the wire is l = 64.714 cm

The current in the wire is I = 8.321 A

The angle is

$\theta=\text{ 19.015}^{\circ}$

To find the magnitude of the force on the wire.

Explanation:

The force can be calculated by the formula

$F=BIlsin\theta$

On substituting the values, the force will be

$\begin{gathered} F=0.075\times8.321\times64.714\times10^{-2}\times sin(19.015)^{\circ} \\ =0.132\text{ N} \end{gathered}$

Thus, the magnitude of the force is 0.132 N