Chapter 20, Problem 3P

A 240-m length of wire stretches between two towers and carries a 120-A current. Determine the magnitude of the force on the wire due to the Earth’s magnetic field of 5.0× 10-5 T which makes an angle of 68⁰ with the wire.

To determine

Force on a straight wire of length 240 m carrying current 120 A and making angle of 68^o with earth’s magnetic field of 5 X 10^-5 T.

Answer to Problem 3P

Solution:

Force = $1.32$ N

Explanation of Solution

Given:

Current I = 120 A; Magnetic field B = 5X 10^-5 T; Length $l$=240m and Ϙ = 68^o

Formula Used:

$F=IlB\sin \theta$

Calculation:

In order to find the value of force on current carrying conductor, we have to use the equation,

$F=IlB\mathrm{Sin}\theta$

= $\frac{120\times 240\times 5\times \sin 68}{100000}$

= $\frac{120\times 240\times 5\times .92}{100000}$

= $1.32\text{ N}$