Problem 2. (4 points) Inducing an emf in a triangular loop inside a solenoid. A

Question

Problem 2. (4 points) Inducing an emf in a triangular loop inside a solenoid. A equilateral triangle, two-turn wire loop l = 1.44 cm on a side is placed inside a solenoid that has a circular cross section of radius r = 3 cm as shown in the end view of Fig. 2. The solenoid is 70 cm long and wound with 100 turns of wire. If the current in the solenoid is 5 A initially and reduced to zero in 4 s, what is the magnitude of the average induced emf in the triangular loop?

Problem 2. (4 points) Inducing an emf in a triangular loop inside a solenoid A equilateral triangle, two-turn wire loop 1-1.44 cm on a side is placed inside a radius r = 3 cm as shown in the end view of Fig. 2. The solenoid is 70 cm long and wound with 100 turns of wire. If the current in the solenoid is 5 A initially and reduced to zero in 4 s, what is the magnitude of the average induced emf in the triangular loop? (A) 20 nV (B) 30 nV (C) 40 nV (D) 68 nV solenoid that has a circular cross section of A' Figure 2. A triangular loop inside a solenoid.

Explanation / Answer

average induced emf is e = rate of change of magnetic flux

e = N*A*(dB/dt)

A = area of the triangular loop = 0.5*b*h = 0.5*(l/2)*h

l^2 = (l/2)^2+h^2

h^2 = l^2-(l/2)^2 = l^2-(l^2/4) = 3*l^2/4

h = sqrt(3)*l/2

A = 0.5*(l/2)*sqrt(3)*l/2 = sqrt(3)*l^2/8 = sqrt(3)*1.44^2*10^-4/8 =0.448*10^-4 m^2

then emf e = N*A*dB/dt

B is the magnetic field inside the solenoid = mu_o*N*I/l = (4*3.142*10^-7*100*5)/0.7 = 8.97*10^-4 T

then

e = N*A*dB/dt = 2*0.448*10^-4*8.97*10^-4/4 = 20*10^-9 = 20 nV

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