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a closed coil having 100 turns is rotated in a uniform magnetic field B=4*10 -4 T about a diameter which is perpendicular to the field. angular velosity of rotation is 300rev per min. the are of the coil is 25cm 2 and its resistance is 4ohm .find (a)the average emfdevelouped in in half a turn from a position ehere the coil is perpendicular to the magnetic field (b)net charge displaced in part (a)


a closed coil having 100 turns is rotated in a uniform magnetic field B=4*10-4T about a diameter which is perpendicular to the field. angular velosity of rotation is 300rev per min. the are of the coil is 25cm2 and its resistance is 4ohm .find


(a)the average emfdevelouped in in half a turn from a position ehere the coil is perpendicular to the magnetic field


(b)net charge displaced in part (a)


Grade:upto college level

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
9 months ago
Φ = magnetic flux
B = magnetic field strength = 4 * 10^-4 Tesla
A = area of circular coil = 25 cm^2 = 0.0025 m^2
ω = angular velocity = 2 π f = 10 π rad/sec
f = frequency (rotations per second ) of the coil rotation = 300/60 = 5 /sec
t = time
N = number of turns in the coil = 100
R = resistance in the circuit of the coil = 4 ohms

Φ = B X A = B * A Cos θ = B * A * Cos ω t = B * A * Cos 2πf t

ε = - N dΦ/dt = N B A (2πf) Sin 2πf t

Emf induced in the coil is = ε = 100 * 4 * 10^-4 * 0.0025 * (10π) Sin (10π t) volts
ε = π Sin (10π t) milli volts. --- equation 1
this is the instantaneous emf induced.

a)
time period for one cycle of rotation = T = 1/f = 1/5 sec = 0.2 sec
time period for 1/2 cycle = 0.1 sec

time average of a sine wave over half cycle of t = 0 sec to 0.1 sec can be obtained by integrating Sin (10πt) dt and dividing by 0.1 sec will be 1/0.1 * 1/(10π) * [- Cos 10π * 0.1 + Cos 10π * 0] = 2/π.
time average (of sin t) wave over half cycle is 2/π. by integration.

Now using the expression from equation 1, the average emf over time over half cycle
= π * (2/π) = 2 milli volts.

b) average emf in time for a full cycle will be zero as the sign of the induced emf will change in the second half cycle. It will cancel the average emf of the first cycle.

c)
charge = current * time = 2 milli volts / 4 ohms * 0.1 secs
= 50 micro coulombs.


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