Chapter 24.2, Problem 22PP

To determine

Strength of magnetic field.

Answer to Problem 22PP

  $B=0.146\text{T}$

Explanation of Solution

Given:

Length of wire, $L=40\text{cm}=0.4\text{m}$

Weight of the wire, $F_g=0.35\text{N}$

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

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:

As force on the wire due to magnetic balances the weight of wire so, $F=F_g=0.35\text{N}$

Also as $F_g$ is downward so $F$ will be upward.

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

On substituting the given values in above relation,

  $\begin{array}{c}0.35=B\times 6\times 0.4\times \sin 90°\\ B=\frac{0.35}{6\times 0.4\times \sin 90°}\\ =0.146\text{T}\end{array}$

Conclusion:

Therefore, the strength of magnetic field is $B=0.146\text{T}$ .