Chapter 24, Problem 69A

To determine

The length of a current-carrying wire kept in a uniform magnetic field and experiencing a given force.

Answer to Problem 69A

The length the current carrying wire is $\text{60 cm}$.

Explanation of Solution

Given:

A wire of a particular length carrying a current of 7.5 A is kept in a uniform magnetic field of strength 0.80 T at an angle of $90°$ with the field. It experiences a force of 3.6 N.

Formula Used:

The force acting on a wire carrying current is dependent on the strength of magnetic field, the current, length of the wire and the angle which the wire makes with the magnetic field. It is mathematically expressed as,

  $F=ILB\left(\sin \theta \right)$

For a given force and the other parameters, the length of the wire can be calculated as,

  $L=\frac{F}{IB\left(\sin \theta \right)}$

Calculation:

For the given strength of magnetic field $B=0.80\text{ T}$ , current $I=7.5\text{ A}$ , force $F=3.6\text{ N}$ and inclination of the wire with the field as $\theta =90°$ , the length of the wire is given by,

  $\begin{array}{c}L=\frac{F}{IB\left(\sin \theta \right)}\\ =\frac{3.6}{7.5\times 0.8\times \left(\sin 90°\right)}\\ =0.6\text{ m}\\ =\text{60 cm}\end{array}$

Conclusion:

The length of the current carrying wire is 60 cm.