6. 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 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Answer:
Maximum emf induced = 0.603 V
Average emf induced = 0 V
Maximum current in the coil = 0.0603 A
Power loss (average) = 0.018 W
(Power which is coming from external rotor)
Circular coil radius, r = 8 cm = 0.08 m
Area of the coil, A = πr2 = π × (0.08)2 m2
Number of turns on the coil, N = 20
Angular speed, ω = 50 rad/s
Strength of magnetic, B = 3 × 10−2 T
Total resistance produced by the loop, R = 10Ω
Maximum emf induced is given as:
e = Nω AB
= 20 × 50 × π × (0.08)2 × 3 × 10−2
= 0.603V
The maximum emf induced in the coil is 0.603 V.
Over a full cycle, the average emf induced in the coil is zero.
Maximum current is given as:
\(I = \frac{e}{R} \)
\(= \frac{0.603}{10} = 0.0603\; A\)
Average power because of the Joule heating:
\(P = \frac{el}{2} \)
\(= \frac{0.603 \times 0.0603}{2} = 0.018\; W\)
The torque produced by the current induced in the coil is opposing the normal rotation of the coil. To keep the rotation of the coil continuously, we must find a source of torque which opposes the torque by the emf, so here the rotor works as an external agent. Hence, dissipated power comes from the external rotor.
7. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb m–2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Answer:
Wire’s Length, l = 10 m
Speed of the wire with which it is falling, v = 5.0 m/s
Strength of magnetic field, B = 0.3 × 10−4 Wb m−2
(a) EMF induced in the wire, e = Blv
= 0.3 × 10−4 × 5 × 10
= 1.5 × 10−3 V
(b) We can determine the direction of the induced current by using Fleming’s right-hand thumb rule, here the current is flowing in the direction from West to East.
(c) In this case, the eastern end of the wire will be having higher potential
8. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Answer:
Current at initial point, I1 : 5.0 A
Current at final pint, I2 : 0.0 A
Therefore, change in current is, dI = I1 – I2 = 5 A
Total time taken, t = 0.1 s
Average EMF, e = 200 V
We have the relation, for self – inductance (L) and average emf of the coil :
\(e = L \frac{di}{dt} \)
\(L = \frac{e}{\left (\frac{di}{dt} \right )} \)
\(= \cfrac{200}{\frac{5}{0.1}} = 4 H\)
Hence, the self – induction of the coil is 4 H.
9. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer:
Given,
Mutual inductance, μ = 1.5H
Current at initial point, I1 : 0 A
Current at final point, I2 : 20 A
Therefore, change in current is, dI = I2 – I1 = 20 – 0 = 20 A
Time taken for the change, t = 0.5s
Emf induced, \(e = \frac{d\phi }{dt}\) ........(1)
dϕ = change in the flux linkage with the coil.
Relation of emf and inductance is:
\(e = \mu \frac{dl}{dt}\) ......(2)
On equating both the equation, we get
\(\frac{d\phi }{dt} = \mu \frac{dl}{dt}\)
\(\frac{d\phi }{dt} = 1.5 \times \left ( 20 \right ) \)
= 30 Wb
Therefore, the change in flux linkage is 30 Wb.
10. A jet plane is travelling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10–4 T and the dip angle is 30°.
Answer:
Speed of the plane with which it is moving, v = 1800 km/h = 500 m/s
Wing span of the jet, l = 25 m
Magnetic field strength by earth, B = 5 × 10−4 T
Dip angle,δ = 30∘
Vertical component of Earth’s magnetic field,
Bv = Bsinδ
= 5 × 10−4 sin30∘
= 2.5 × 10−4 T
Difference in voltage between both the ends can be calculated as:
e = (Bv) × l × v
= 2.5 × 10−4 × 25 × 500
= 3.125V
Hence, the voltage difference developed between the ends of the wings is 3.125 V.
11. Let us assume that the loop in question number 4 is stationary or constant but the current source which is feeding the electromagnet which is producing the magnetic field is slowly decreased. It was having an initial value of 0.3 T and the rate of reducing the field is 0.02 T/sec. If the cut is joined to form the loop having a resistance of 1.6Ω. Calculate how much power is lost in the form of heat? What is the source of this power?
Answer:
Rectangular loop is having sides as 8 cm and 2 cm.
Therefore, the area of the loop will be, A = L × B
= 8 cm × 2 cm
= 16cm2
= 16 × 10−4 cm2
Value of magnetic field at initial phase, B’ = 0.3 T
Magnetic fields decreasing rate, \(\frac{dB}{dt} = 0.02 T/s\)
Emf induced in the loop is:
\(e = \frac{d\phi }{dt}\)
dϕ = change in flux in the loop area
= AB
\(∴ e = \frac{d\left ( AB \right )}{dt} = \frac{AdB}{dt}\)
= 16 × 10−4 × 0.02 = 0.32 × 10−4 V
Resistance in the loop will be, R = 1.6Ω
The current developed in the loop will be:
\(i = \frac{e}{R} \)
\(= \frac{032 \times 10 ^{-4}}{1.6} = 2 \times 10^{-5} A\)
Power loss in the loop in the form of the heat is:
P = i2R
= (2 × 10−5)2 × 1.6
= 6.4 × 10−10 W
An external agent is a source for this heat loss, which is responsible for the change in the magnetic field with time.