Chapter 24, Problem 96A

To determine

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

Answer to Problem 96A

The force on the wire is $5.52\text{ N}$

Explanation of Solution

Given:

A wire, 2.5 cm in diameter is wound around cylindrical form making 250 loops. The resistance of the wire is $8.0\Omega$ and is fed with a voltage of 15 V. It is kept in a uniform magnetic field of strength 0.15 T.The arrangement of the wire indicates 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,

  $F=ILB\left(\sin \theta \right)$

Here, the length of the wire can be calculated by considering the number of loops and the circumference of each loop.

Since the wire is wound around a cylinder, the length of one loop is its circumference given by,

  $C=2\pi r$

Now, the length of the whole wire consisting N loops is given by,

  $L=NC$

To find the current through the wire, Ohm’s law can be used for a given applied voltage and resistance of the wire as,

  $I=\frac{V}{R}$

Calculation:

The length of one loop which is the circumference, is given by,

  $\begin{array}{c}C=2\times 3.14\times \frac{2.5}{2}\times {10}^{-2}\\ =7.85\text{ cm}\end{array}$

Now, the length of the whole wire is given by,

  $\begin{array}{c}L=250\times 7.85\times {10}^{-2}\\ =19.62\text{ m}\end{array}$

The current through the wire for the supply voltage V =15 V and the resistance $R = 8.0\Omega$ can be calculated as,

  $\begin{array}{c}I=\frac{15}{8}\\ =1.875\text{ A}\end{array}$

For the given strength of magnetic field $B=0.1\text{5 T}$ , current $I=1.875\text{ A}$ and length $L=19.62\text{ m}$ , the force at angle $\theta =90°$ is given by,

  $\begin{array}{c}F=1.875\times 19.62\times 0.15\times \left(\sin 90°\right)\\ =5.52\text{ N}\end{array}$

Conclusion:

The force on the current carrying wire is 5.52 N.