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lakkis [162]
2 years ago
5

The picture shows a loop of wire rotating in a magnetic field in a generator. Determine the direction of the current through the

load as the loop rotates from 180 to 270 degrees as shown in the picture.
A) From right to left

B) From left to right

C) There is not enough information to tell which direction the current will be in the circuit.

Physics
2 answers:
BaLLatris [955]2 years ago
3 0
I believe the answer is b
Fiesta28 [93]2 years ago
3 0

Answer:

B) From left to right

Explanation:

As per Lenz law the direction of induced current is always in such a way that it will oppose the causes due to which it is induced in a closed loop

Like if the flux is increasing in the loop then induced current is in such a way that it will try to decrease the flux

If flux is decreasing with time in a closed loop then induced current will flow in such a way that it will try to increase the decreasing flux

So in all the way it will always oppose the changes

so here while loop is moving from 180 degree to 270 degree then the flux is decreasing so the direction of induced current in the load will be from left to right

so here correct answer will be

B) From left to right

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21.2\times 10^{-6} T

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B = magnitude of magnetic field by wires at A and C

B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )

B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )

B = 10\times 10^{-6} T

B' = magnitude of magnetic field by wire at B

B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )

B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )

B' = 7.02\times 10^{-6} T

Net magnitude of the magnetic field at D is given as

B_{net} = \sqrt{B^{2}+B^{2}} + B'

B_{net} = \sqrt{2} B + B'

B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})

B_{net} = 21.2\times 10^{-6} T

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