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 4.00 cm and that of the lower circle is 8.00 cm. The wire has a uniform resistance per unit length of 2.00 ohm/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 1.70 T/s. (a) Find the magnitude of the induced current in the wire. (b)Find the direction of the induced current in the wire. (Select all that apply.)

Explanation / Answer

E = A * dB/dt
upper loop:
r = 4 cm = 0.04 m
area = pi*r^2 = pi* (0.04)^2 = 5.03*10^-3 m^2
induced emf = A*dB/dt = 5.03*10^-3 *1.7 = 0.00855
total length of wire = 2*pi*r = 2*pi*0.04 = 0.25 m
Resistance = 0.25*2 = 0.5 ohm

bottom loop:
r = 8 cm = 0.08 m
area = pi*r^2 = pi* (0.08)^2 = 0.02 m^2
induced emf = A*dB/dt = 0.02 *1.7 = 0.0342 V
total length of wire = 2*pi*r = 2*pi*0.08 = 0.5 m
Resistance = 0.5*2 = 1 ohm

a)
potential difference = 0.0342 - 0.00855 = 0.026 V
toatl resistance = 0.5+1 = 1.5 ohm
Current = 0.026/1.5 = 0.017 A

b)
Since voltage induced ios more is lower loop, use the lenz'a law to know the direction of induced voltage in lower loop. same will give the current direction.
In bottom loop : counterclockwise
In upper loop it will be opposite.
in upper loop :clockwise
option 1 and 4

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