Question

Part A

Determine the magnitude and direction of the force between two parallel wires 30 m long and 3.0 cm   apart, each carrying 30 A in the same direction.

The force on each wire will be attractive.

The force on each wire will be repulsive.

The wires will not interact.

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A long horizontal wire carries 23.0 A of current due north. The Earth's field 20.0 cm due west of the wire points north but downward, 45 ∘ below the horizontal, and has magnitude of 5.0×10−5T.

Part A

What is the magnitude of the net magnetic field 20.0 cm due west of the wire?

Part B

What is the angle below the horizontal of the net magnetic field 20.0 cm due west of the wire?

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A

Determine the magnetic field midway between two long straight wires 2.0 cm apart in terms of the current I in one when the other carries 25 A. Assume these currents are in the same direction.

Bnet=(2.0×10−5T/A)(I+25A)

Bnet=(1.0×10−5T/A)(I−25A)

Bnet=(1.0×10−5T/A)(I+25A)

Bnet=(2.0×10−5T/A)(I−25A)

B

Determine the magnetic field midway between two long straight wires 2.0 cm apart in terms of the current I in one when the other carries 25 A. Assume these currents are in opposite directions.

Bnet=(2.0×10−5T/A)(I−25A)

Bnet=(1.0×10−5T/A)(I+25A)

Bnet=(1.0×10−5T/A)(I−25A)

Bnet=(2.0×10−5T/A)(I+25A)

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A 33.0-cm long solenoid 1.25 cm in diameter is to produce a field of 4.45 mT at its center.

Part A

How much current should the solenoid carry if it has 985 turns of the wire?

Answer 1

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