A certain length of wire carrying a current of 10.0 A is oriented horizontally i

Question

A certain length of wire carrying a current of 10.0 A is oriented horizontally in a region where the earth's magnetic field of 5.00 x 10^-5 Tesla had a dip angle of 46.5 degrees. If the magnetic force on the wire is 1.57 x 10^-4N, what is the length of the wire? A certain length of wire carrying a current of 10.0 A is oriented horizontally in a region where the earth's magnetic field of 5.00 x 10^-5 Tesla had a dip angle of 46.5 degrees. If the magnetic force on the wire is 1.57 x 10^-4N, what is the length of the wire?

Explanation / Answer

Here ,

I = 10 A

B = 5 *10^-5 T

theta = 46.5 degree

let the length of the wire is L

F = B*I*L * sin(theta)

1.57 *10^-4 = 5 *10^-5 * 10 * L * sin(46.5 degree)

solving for L

L = 0.433 m

the length of the wire is 0.433 m

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