Chapter 24.2, Problem 21PP

To determine

Strength of magnetic field.

Answer to Problem 21PP

  $B=0.133\text{T}$

Explanation of Solution

Given:

Length of wire, $L=75\text{cm}=0.75\text{m}$

Force on the wire, $F=0.6\text{N}$

Current through the wire, $I=6\text{A}$

Angle of wire with magnetic field, $\theta =90°$

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,

  $\begin{array}{c}0.6=B\times 6\times 0.75\times \sin 90°\\ B=\frac{0.6}{6\times 0.75\times \sin 90°}\\ =0.133\text{T}\end{array}$

Conclusion:

Therefore, the strength of magnetic field is $B=0.133\text{T}$ .