The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of

Question

3 years ago

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The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of

0.10Ω, and it enters the field at t = 0s. Find the induced current in the loop as a function of time. Give your answer as a graph of I versus t from t = 0s to t = 0.020s.

Physics

2 answers:

PolarNik [594] · Answer 1 · 2022-05-31T14:51:24+03:00

The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions

Sati [7] · Answer 2 · 2022-05-31T15:55:59+03:00

The induced current in the loop is 20 A

<h3>Further explanation </h3>

Lenz's law said that when EMF is generated by the change in magnetic flux the polarity of the induced EMF produces an induced current. It induced electric current flows in a direction such that the current opposes the change that induced it

The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of 0.10Ω, and it enters the field at t = 0s. Find the induced current in the loop as a function of time. Give your answer as a graph of I versus t from t = 0s to t = 0.020s.

B = 0.8 T

L = 5 cm

v = 10 m/s

R = 0.10 Ω

I = ?

∈=E

Ф=o

$I = \frac{E }{R} = \frac{1}{R} |\frac{d o_{m}}{dt} | = \frac{1}{R} \frac{d}{dt} |B.dA|\\I = \frac{1}{R} \frac{d}{dt} |B.dl\Delta s|\\I = \frac{B*l}{R} \frac{ds}{dt} \\I = \frac{B*lv}{R} = \frac{0.80 T*0.05m*50m/s}{0.10 ohm}$