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A long straight wire carries a current I 1 = 3.9 A and is a distance h = 18.4 cm

ID: 2280013 • Letter: A

Question

A long straight wire carries a current I1 = 3.9 A and is a distance h = 18.4 cm from a rectangular current loop that carries a current I2= 1.9 A. If the loop has a length L = 49 cm and a width w = 24 cm, what is the magnitude of the force exerted by the wire on the loop?



=______N



What is the force exerted by the current loop on the wire?

=______N


A long straight wire carries a current I1 = 3.9 A and is a distance h = 18.4 cm from a rectangular current loop that carries a current I2= 1.9 A. If the loop has a length L = 49 cm and a width w = 24 cm, what is the magnitude of the force exerted by the wire on the loop?

Explanation / Answer

The long straight wire produces a magnetic field of
B = u0 I1 / (2 pi r)
B = (2*10^-7 * 3.9) / r

Force on each parallel side of the loop :
F = I2 L B

Nearer side of the loop is at a distance r = h

This gives a force :
F1 = (1.9 * 0.49 * 2*10^-7 * 3.9) / 0.184
F1 = 3.947 * 10^-6 N

Farther side of the loop is at a distance :
r = h + w
r = 0.146 + 0.24
r = 0.424

This gives a force of F2 :
= 1.5 * 0.49 * 2*10^-7 * 3.9 / 0.424
= 1.352 * 10^-6 N

Since both forces are in opposite directions so subtract.

The forces on the sides of the loop perpendicular to the long wire are also in opposite directions, so they cancel each other completely.

= (3.946*10^-6) - (1.352*10^-6)
= 2.595 * 10^-6 N towards the straight wire

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