Complete the 3 questions below. The loop in the image is moving through the magn

Question

Complete the 3 questions below. The loop in the image is moving through the magnetic field, of strength 0.200 T, at a speed of 53.0 m/s. If the resistance in the loop is 1.20 Ohm and the width of the wire is x = 10.0 cm, what is the magnitude and direction of the induced current? 0.683 A, counterclockwise. 3.53 A, dock wise. 3.53 A, counterclockwise. 0.883 A, clockwise. A magnetic field is expressed with the equation B = 6.00 - 6.00t^2, A square loop is perpendicular to the magnetic field with dimensions 11.0 cm times 11.0 cm, and a resistance of 0.180 Ohm. What is the induced current at time t = 9.90 s? 0.403 A -7.58 A -35.5 A 1.36 A, What is the magnetic x rough the loop in the figure? B = 2.70 T, a = 20 cm, b = 7.00 cm, r_1 = 3.90 cm and r_2 = 4.10 cm. 1.20 times 10^-5 Tm^2 0.00227 Tm^2 1.08 times 10^-5 Tm^2 -0.00227 Tm^2

Explanation / Answer

Induced emf is e = B*l*v

B = 0.2 T ,l = 10 cm = 0.1 m and v = 53 m/sec

then e = B*l*v = 0.2*0.1*53 = 1.06 V

induced current is i = e/R = 1.06/1.2 =0.883 A

according to Lenz's law the direction of current is counterclockwise

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2) induced current i = e/R

induced emf is e = d(phi)/dt = A*(dB/dt) = 0.11*0.11*(6-(2*6*t)) = 0.11^2*(6-12*t)

at t = 9.9 sec

then

e = 0.11^2*(6-(12*9.9)) = -1.36 V

i = e/R = -1.36/0.18 =-7.58 A

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