A piece of insulated wire is shaped into a figure eight as shown in the figure b

Question

A piece of insulated wire is shaped into a figure eight as shown in the figure below. For simplicity, model the two halves of the figure eight as circles. The radius of the upper circle is 3.00 cm and that of the lower circle is 8.00 cm. The wire has a uniform resistance per unit length of 4.00 /m. A uniform magnetic field is applied perpendicular to the plane of the two circles, in the direction shown. The magnetic field is increasing at a constant rate of 2.60 T/s xXxxX xXXXx (a) Find the magnitude of the induced current in the wire. 0.03043 Your response differs from the correct answer by more than 10%. Double check your calculations, A (b) Find the direction of the induced current in the wire. (Select all that apply.) clockwise in the upper loop clockwise in the lower loop counterclockwise in the upper loop | counterclockwise in the lower loop

Explanation / Answer

B = A*B = (rs^2 rb^2)B

So using Ampere’s law

= - (dB/dt)

*(rb^2 rs^2)*dB/dt

The resistance of the entire figure eight is

R = (2rs +2rb)

And plugging that into Ohm’s law

= V = IR

I = (rb^2 rs^2)*dB/dt / 2(rs + rb)

I = (0.08^2 0.03^2)*4.00 / 2*2.60(0.03 + 0.08)

I = 0.0384 = 38.4mA

Pick a direction for the current to be counterclockwise in the bottom loop (so clockwise in the top). Thus, the area vector of the top loop is antiparallel to B and that of the bottom loop is parallel to B

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