Chapter 24, Problem 68A

To determine

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

Answer to Problem 68A

The force on the wire is $\text{0 N}$

Explanation of Solution

Given:

A wire of length 35 cm carrying a current of 4.5 A is kept in a uniform magnetic field of strength 0.53 T in a direction parallel to the field.

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,

  $F=ILB\left(\sin \theta \right)$

Calculation:

The current carrying wire is kept in a direction parallel to the field. This implies an inclination of $\theta =0°$

For the given strength of magnetic field $B=0.53\text{ T}$ , current $I=4.5\text{ A}$ and length $L=35\text{ cm}$ , the force at angle $\theta =0°$ is given by,

  $\begin{array}{c}F=4.5\times 0.35\times 0.53\times \left(\sin 0°\right)\\ =0\text{ N}\end{array}$

This implies that, whenever a current carrying wire is kept in a direction parallel to the magnetic field, it does not experience any force.

Conclusion:

The force on the current carrying wire is 0 N.