Question

12.The figure shows the cross section of a solenoid with 2 x 104 loops per m, (N/R 2x104 m-1), with a radius of 6 cm. Inside this solenoid is a circular coil of wire consisting of 45 loops of wire with radius 4.0 cm whose total resistance is 6 ?. The current in the solenoid changes in time and is l(t) lo cos (o t), with 10-2.5 Amps, ?-400 rad/s (please note the radians) and t is measured in seconds. What is the instantaneous power dissipated in the circular coil at time t 2 ms? solenoid

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Answer #1

There will be induced current inside the wire

i=\dfrac{E}{R}=\dfrac{\dfrac{d(BA)}{dt}}{R}

i=\dfrac{\dfrac{d(\mu n i\pi r^2)}{dt}}{R}

i=\dfrac{\mu n \omega i_o sin\omega t}{R}

Using all the given values,

i=\dfrac{4\pi*10^{-7} 2*10^4*400*2.5 sin(400*2*10^{-3}) \pi0.04^2 *45}{6}

i=0.68A

Power dissipated,

P=i^2R

P=0.68^2*6=2.77watt

Thankyou

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