Chapter 24, Problem 67A

To determine

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

Answer to Problem 67A

The force on the wire is $\text{0}\text{.45 N}$

Explanation of Solution

Given:

A wire of length 25 cm carrying a current of 6.0 A is kept in a uniform magnetic field of strength 0.30 T at an angle of $90°$ with 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:

For the given strength of magnetic field $B=0.30\text{ T}$ , current $I=6.0\text{ A}$ and length $L=25\text{ cm}$ , the force at angle $\theta =90°$ is given by,

  $\begin{array}{c}F=6.0\times 0.25\times 0.3\times \left(\sin 90°\right)\\ =0.45\text{ N}\end{array}$

Conclusion:

The force on the current carrying wire is 0.45 N.