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  • A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s ^-1 in a uniform horizontal magnetic field of magnitude 3 .0 × 10^-2T.

Q6.6     A circular coil of radius 8.0 cm and 20 turns  is rotated about its vertical diameter with an angular speed of 50 \: rads\: s^{-1}  in a uniform horizontal magnetic field of magnitude 3.0\times 10^{-2} T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10\Omega , calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Answers (1)

best_answer

Given

The radius of the circular loop r=8cm=0.08m

Number of turns N = 20

Flux through each turn 

\phi = B.A=BAcos\theta =BAcos\omega t=B\pi r^2cos\omega t

Flux through N turn

\phi =NB\pi r^2cos\omega t

Induce emf:

e=\frac{d\phi }{dt}=\frac{d(NB\pi r^2cos\omega t)}{dt}=NBr^2\pi \omega sin\omega

Now,

maximum induced emf (when sin function will be maximum)

 e_{max}=NB\pi r^2\omega=20*50*\pi *(0.08)^2*3*10^{-2}=0.603V

 

Average induced emf

e_{average}=0  as the average value of sin function is zero,

 

Maximum current when resistance R of the loop is 10\Omega.

I_{max}=\frac{e_{max}}{R}=\frac{0.603}{10}=0.0603A

 

Power loss :

P_{loss}= \frac{1}{2}E_0I_0=\frac{1}{2}(0.603)(0.0603)=0.018W

Here, power is getting lost as emf is induced and emf is inducing because we are MOVING the conductor in the magnetic field. Hence external force through which we are rotating is the source of this power.

 

Posted by

Pankaj Sanodiya

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