If the EMF produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amp·meter at a speed of 4.20 meters/second, what is the length of the wire in the magnetic field?

a. 0.27 meters

b. 2.8 meters

c. 0.020 meters

d. 3.7 meters

Question

Mariah Solis

Chemistry

4 May, 08:01

If the EMF produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amp·meter at a speed of 4.20 meters/second, what is the length of the wire in the magnetic field?

a. 0.27 meters

b. 2.8 meters

c. 0.020 meters

d. 3.7 meters

+5

Answers (2)

Know the Answer?

Melany Blevins · Answer 1

Melany Blevins 4 May, 08:09

0

The answer to your question is B

Bob · Answer 2

B = 0.075 N/amp*m

V = 4.20 m/s

EMF = 0.88 Volts

L = ?

Formula = EMF = L x B x V

Since we have EMF and we’re looking for L, the formula will now be:

L = EMF x B x V

Solution:

= 0.88 x 0.075 x 4.2

= 2.8

The length of the wire in the magnetic field is 2.8m

The answer is letter B.

If the EMF produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amp·meter at a speed of 4.20 meters/second, what is the length of the wire in the magnetic field?a. 0.27 metersb. 2.8 metersc. 0.020 metersd. 3.7 meters

If the EMF produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amp·meter at a speed of 4.20 meters/second, what is the length of the wire in the magnetic field?

a. 0.27 meters

b. 2.8 meters

c. 0.020 meters

d. 3.7 meters