Chapter 30, Problem 028 Partially correct answer. Your answer is partially corre

Question

Chapter 30, Problem 028 Partially correct answer. Your answer is partially correct. Try again. In the figure, a rectangular loop of wire with length a = 2.3 cm, width b = 1.5 cm, and resistance R = 0.90 m is placed near an infinitely long wire carrying current i = 6.5 A. The loop is then moved away from the wire at a constant speed v = 5.6 mm/s. When the center of the loop is at distance r = 2.5 cm, what are (a) the magnitude of the magnetic flux through the loop and (b) the current in amperes induced in the loop? (a) Number Entry field with incorrect answer 1.8e-8 Units Entry field with correct answer (b) Number Entry field with incorrect answer 7.44e-6 Units Entry field with correct answer

Explanation / Answer

given that

length a = 2.3 cm = 0.023 m

width b = 1.5 cm = 0.015 m

R = 0.90 m ohm = 0.90*10^(-3) ohm

i = 6.5 A

v = 5.6 mm/s = 5.6 *10^(-3) m/s

we know that

megnetic field at distance r from wire is

B = u0* i / (2* pi* r)

so,

flux = integration (B.da) = int (u0* i / (2*pi * x) a. dx) from r - b/2 to r + b/2

flux = u0* i * a / 2*pi (ln(r + b/2) - ln(r - b/2))                   .......eq1

put all the values in eq1

r = 2.5 cm = 0.025 m

flux = 4*pi*10^(-7) * 6.5 * 0.023 / 2*pi (ln(0.025+.0075) - ln(.025-.0075)

flux = 0.299*10^(-7) / 0.62

flux = 0.482*10^(-7) weber

flux = 4.82*10^(-8) weber

we know that

current in loop

i' = emf / R

and emf = - rate of change of flux

so,

i' = u0* i * a * (b/(r^2 - b^2/4)) * v / (2*pi* R)

i' = (4*pi*10^(-7) * 6.5 * 0.023 * (0.015 / (0.025)^2 - (0.015)^2/4)) * 5.6 *10^(-3)) / 2 * pi *0.90 *10^(-3)

i' = 2 * 10^(-7) * 6.5 * 0.023 * 26.37 * 5.6 / 0.90

i' = 44.16*10^(-7) / 0.90

i' = 49.06*10^(-7) A

i' = 0.49*10^(-6) A

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