11. Two parallel wires have the same current of 30 A in the opposite directions. If the force per unit length (F/L) is 2.52 x 10 N/m, how close to each other would they be? a. 7.14 cm b. 71.4 m c. 0.714 cm d. none of these

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Chapter31: Faraday’s Law
Section: Chapter Questions
Problem 31.42P: 100-turn square coil of side 20.0 cm rotates about a vertical axis at 1.50 103 rev/min as indicated...
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Question 11

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9. A multi-loop coil has 2.00 A current passing through and its radius is 30 cm. If the B field at the center of
the coil is 0.0017 T, find the number of loops the coil has.
a. 1275
b. 203
c. 812
d. 406
10. A solenoid has 2500 turns over the length of 70 cm. How much current is needed to generate 0.50 T
magnetic field at the center of the solenoid?
a. 11.4 A
b. 11141 A
c. 111.4 A
d. none of these
11. Two parallel wires have the same current of 30 A in the opposite directions. If the force per unit length
(F/L) is 2.52 x 10 N/m, how close to each other would they be?
a. 7.14 cm
b. 71.4 m
c. 0.714 cm
d. none of these
The simple design with a slidable wire section can be treated as a generator. If the magnetic field is 0.65 T
and the speed of the slide wire is 4.5 m/sec, (L=75 cm)
Transcribed Image Text:28% O 10:21 PM it 9. A multi-loop coil has 2.00 A current passing through and its radius is 30 cm. If the B field at the center of the coil is 0.0017 T, find the number of loops the coil has. a. 1275 b. 203 c. 812 d. 406 10. A solenoid has 2500 turns over the length of 70 cm. How much current is needed to generate 0.50 T magnetic field at the center of the solenoid? a. 11.4 A b. 11141 A c. 111.4 A d. none of these 11. Two parallel wires have the same current of 30 A in the opposite directions. If the force per unit length (F/L) is 2.52 x 10 N/m, how close to each other would they be? a. 7.14 cm b. 71.4 m c. 0.714 cm d. none of these The simple design with a slidable wire section can be treated as a generator. If the magnetic field is 0.65 T and the speed of the slide wire is 4.5 m/sec, (L=75 cm)
Expert Solution
Step 1

11) Force per unit length of wire = μo I1I2/2πd2.52 * 10-4 = 4π * 10-7 *30* 30/(2πd)d = 0.714 m d = 71.4 cm 

 

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