Two insulated wires, each 2.50 m long, are taped together to form a two-wire unit that is 2.50 m long. One wire carries a current of 7.00 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 79.0° relative to a magnetic field whose magnitude is 0.360 T. The magnitude of the net magnetic force experienced by the two-wire unit is 3.13 N. What is the current I?

Question

Theresa Gay

Physics

21 July, 17:55

Two insulated wires, each 2.50 m long, are taped together to form a two-wire unit that is 2.50 m long. One wire carries a current of 7.00 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 79.0° relative to a magnetic field whose magnitude is 0.360 T. The magnitude of the net magnetic force experienced by the two-wire unit is 3.13 N. What is the current I?

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Answers (1)

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Madilyn Hanna · Answer 1

3.46 A

Explanation:

The force (F) exerted on a wire of a particular length (L) carrying current (I) through a magnetic field (B) at an angle (θ) to the magnetic field is given as

F = (B) (I) (L) sin θ

F = 3.13 N

B = 0.360 T

I = ?

L = 2.50 m

θ = 79°

3.13 = (0.360 * I * 2.5 * sin 79°)

0.8835 I = 3.13

I = 3.54 A

But this is the resultant current in this magnetic field.

Since the two wires are conducting current in opposite directions,

Resultant current = 7 - (current in the other wire)

Current in the other wire = 7 - 3.54 = 3.46 A