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A loop is pulled with a force F to the right to maintain a constant speed of 8.0 m/s. The loop has a length of 0.15m, a width of 0.08m, and a resistance of 200.0. The magnitude of the field Is 1.2T. (a) what is the magnitude of the emf induced in the loop? (b) what is the induced current in the loop? (c) determine the magnitude of the force F requires to pull the loop.

Respuesta :

Answer:

Explanation:

The emf induced in a straight conductor of length l moving with velocity v perpendicular to a magnetic field B is

E = Blv

where B, l and v are perpendicular. The emf, E is in volts

Magnetic field, B is in webers/m2

= 1.2 T

length is in meters

= 0.15 m

velocity, v is in m/sec

= 8 m/s

E = 1.2 × 0.15 × 8

= 1.44 V

B.

Using ohm's law,

V = I × R

I = 1.44/200

= 0.0072 A

C.

F = B × I × L

= 1.2 × 0.0072 × 0.15

= 0.001296 N

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