Question

You find a square loop of wire (m = 7.5 grams) that has a side length of 5cm and a total resistance of 5mΩ. It is sitting halfway on a table that has a B-field which is perpendicular to the plane of the table. You find that you can adjust the value of the B-field using a knob on the table. At t = 0s the B-field increases in size from 0T to 2T over 0.015s.

a) What is the max induced current in the loop?

b) The loop will feel a force that will quickly “push” the loop, what is the speed gained by the loop during this quick “push”?

*You can assume that the loop doesn’t move a measurable amount during 0.015s the B-filed increases.

Answer 1

From Faraday's law the induced emf = - d(phi)/dt
phi is magnetic flux , phi = B*A* cos theta

here theta = 0 degrees

length of the square loop l = 5 cm , area A = L^2 = 5*5*10^-4 m^2 = 25*10^-4 m^2
given half of the loop is inside the magnetic field , so the effective area is A1 = 12.5*10^-4 m^2
Resistance R = 5 m.ohm

the change in magnetic field is dB = 2 T , during the time dt = 0.015 s
a)

   so induced emf , e = -12.5*10^-4*2/0.015 V

   e = 0.1667 V

From Ohm's law e = i*R

   i = e/R = 0.1667/(5*10^-3) A = 33.34 A
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b)

And from the concept of induced emf , e = B*v*L sin theta , theta = 90 degrees

   v = e/B*L
   v = 0.1667/(2*2.5*10^-2) m/s

   v = 3.334 m/s

the speed gained by the loop during this quick “push" is v = 3.334 m/s