Chapter 24.2, Problem 20PP

To determine

Force on the wire.

Answer to Problem 20PP

  $F=1.6\text{N}$

Explanation of Solution

Given:

Length of wire, $L=0.5\text{m}$

Current through the wire, $I=8\text{A}$

Angle of wire with magnetic field, $\theta ={90}^o$

Strength of magnetic field, $B=0.4\text{T}$

Formula Used:

Force on current carrying wire in magnetic field is given as,

  $F=BIL\sin \theta$

Where, $B$ is strength of magnetic field, $I$ is current through the wire, $L$ is length of wire and $\theta$ is angle of wire with magnetic field.

Calculation:

On substituting the given values in above relation,

  $\begin{array}{c}F=0.4\times 8\times 0.5\times \sin 90°\\ =1.6\times 1\\ =1.6\text{N}\end{array}$

Conclusion:

Therefore, the force acting on the wire is $F=1.6\text{N}$ .