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A uniform magnetic field vector B is established perpendicular to the plane of a loop of radius 2 cm, resistance 0.8 , and negligible self-inductance. The magnitude of vector B is increasing at a rate of 40 mT/s. Find the following values.

(a) the induced emf in the loop _______mV.
(b) the induced current in the loop________ mA.
(c) the rate of Joule heating in the loop_________ µW.

1 Answer

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Answer

the expression of induced emf


e = (d)/(dt)(nAB cos\theta)

θ is the angle between the plane and magnetic force


e = nA cos\theta(d)/(dt)(B )

θ = 0°

a)
e = nA cos 0^0(d)/(dt)(B )


e = nA (dB)/(dt)


e = 1 * \pi \0.02^2 * 40 * 10^(-3)


e = 1 * \pi \0.02^2 * 40 * 10^(-3)

e = 0.05 mV

b) induced current


i = (e)/(R)


i = (0.05* 10^(-3))/(0.8)

i = 0.063 mA

c) Rate of heat change

P = I² R

P = 0.063² x 0.8

P = 0.003175 µW

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