Electromagnetic Induction & Alternating Currents: Important Questions

Very Short Answers [1 Marks]

Ques. An infinitely-long current-carrying wire is passing through the center of a circular wire loop. If the current through the straight wire is decreasing at a steady rate, then what will be the direction of induced current in the loop if any. (NCERT) 

Ans. Magnetic lines of forces lie in the plane of the loop, so there will be no current induced in the loop.

Ques. Two bar magnets are shifted towards a metallic loop connected across a capacitor ‘C’ as illustrated. Predict the polarity of the capacitor. (All India 2013)

Ans. A side of the Cl capacitor will be positively charged and the lower plate B is negative when both the magnets move towards the loop. Thus it stimulates the induced current in a clockwise position.

Ques. Elaborate Lenz’s law.

Ans. In electromagnetism, Lenz’s law claims that “the polarity of induced EMF is such that it tends to produce a current, which opposes the change in magnetic flux that induced it”.

Ques. A plot of magnetic flux (\(\emptyset\)) versus current (I) is shown in the figure for two inductors A and B. Which of the two has a larger value of self-inductance? (Delhi, 2010)

Ans.\(\emptyset\) = LI

L = \(\emptyset\)I = slope

The slope of A is < slope of B

∴ Inductor A has a larger measure of self-inductance than inductor B.Ques. Define the self-inductance of a coil. Write its S.I. unit. (All India 2010) 

Ans. Self-inductance can be described as the tendency of a coil, by virtue, to withstand changes in current flowing through it. Henry (H) is known to be its S.I unit.
Ques. A bar magnet is moved in the directed way the arrow shows between two coils, PQ and CD. as per the image displayed, determine the directions of induced current in each coil. (All India 2012) 

Ans. Lenz’s law points out that the ends of both the coils come close to the magnet and behave as a south pole. So, the current induced in both the coils will flow clockwise when seen from the magnet side.

Ques. A copper plate’s motion is damped during oscillation between two poles of a magnet. Determine the cause of the damping. (All India 2013) 

Ans. Eddy current is what causes the damping.
 

Also Read:

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Short Answers [2 Marks]

Ques. A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give a reason. (All India 2013)

Ans. The reason states that because of the eddy current, the upper force of the core of the electromagnet gets the north polarity. According to Lenz’s law, the lower face of the disc will also get the north polarity. Because of the force of the repulsion, between the lower face ( north pole ) of the core electromagnet, the disc jumps up to a positive height.

Ques. Two identical loops, one of copper and the other of aluminum are rotated with the same angular speed in the same magnetic field. Compare,

(1)The induced EMF

(2)The current produced in the two coils. (All India 2010)

Ans. 

1. The Induced EMF in a coil, ε = NBA ω sin ωt
   Induced EMF will be the same in both the loops as the angular speed is the same.
2. The current induced in a loop =
   I= εR = εAρl
   (More current is induced as the resistivity of copper becomes lesser.)

Ques. The figure given shows an arrangement by which current flows through the bulb (X) connected with coil B when a.c. is passed through coil A.

(1)Name the phenomenon involved.

(2)If a copper sheet is inserted in the gap between the coils, explain how the brightness of the bulb would change. (All India 2008)

Ans. 

1. Mutual induction is the phenomenon that is involved.
2. When the copper sheet is inserted, eddy currents are set up in it which react to the passage of magnetic fluid. The induced EMF in coil B decreases and it dims the brightness of the bulb.

Ques. Highlight the working mechanism of a transformer. Can a transformer be used to step up or step down a D.C voltage? Give justifiable reasons. (All India 2009)

Ans. Transformers follow the rules of mutual induction i.e. when a changing current is passed through one of the two inductively coupled coils, an induced EMF is set up in the other coil. No, the transformer cannot be used to increase or reduce the D.C voltage because A.C voltage cannot produce a change in magnetic flux.

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Long Answers [3 Marks]

Ques. A coil Q is connected to low voltage bulb B and placed near another coil P as shown in the figure. Justify why the Bulb gets dimmer if the coil Q is moved towards the left. (Delhi 2010)

Ans. When coil Q travels towards the left, the magnetic flux which is linked with coil Q decreases and sometimes even reduces to zero at some distance. The EMF induced may decrease and bulb B eventually gets subdued.

Ques. The current through two inductors of self-inductance 15 mH and 25 mH is increasing with time at the same rate. Draw graphs showing the variation of the

(a)EMF induced with the rate of change of current

(b)Energy stored in each inductor with the current flowing through it.

(c)The energy stored in the coils, if the power dissipated in the coils is the same. (All India 2017) 

Ans. Let's follow the process with L1 = 15 mH and L2 = 25 mH

Also Check: Difference between AC and DC Generator

Ques. How is the mutual inductance of a pair of coils affected when

(1)Separation between the coils is increased.

(2)A thin iron sheet is placed between two coils, other factors remaining the same. Explain the answer in each case.

Ans.

1. The flux connected with the secondary coils decreases when the separation between the coils increases. So, the mutual induction decreases.
2. As mutual inductance, M∝μr is (Relative permeability of material). So according to the theory, mutual induction will increase.

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Very Long Answers [5 Marks]

Ques. A jet plane is traveling towards the west at a speed of 1800 km/hr. What is the voltage difference developed between the ends of the wing having a span of 25m, if the earth’s magnetic field at the location has a magnitude of 5 x 10^-4 T and the dip angle is 30°.

Ans. The speed of the jet plane, v = 1800km/h = 500 m/

Wingspan of jet plane, l = 25m

Earth’s magnetic field strength, B=5×10−4T

Angle of dip, \(\delta\) = 30°

Considering the vertical component of EMF

Bv = B sin\(\delta\)

⇒ Bv = 5×10^-4 sin 30⁰

⇒ 2.5×10^-4 T

Therefore, the voltage difference between the wings' end

E = (Bv)×l×v

⇒ e=2.5×10^-4 ×25×500

= 3.125V

So, the answer to the voltage difference developed between the ends of the wings is 3.125v

Ques. Prove that an ideal capacitor in an A.C circuit does not dissipate power. (Delhi 2008) 

Ans. When an A.C is used in a capacitor, the current leads the voltage in phase by 2 radian because average power is associated with a capacitor.

V = V₀ sin \(\omega\)t, I = I₀ sin (\(\omega\) + \(\pi \over 2\) ) – I0 cos \(\omega t\)

Work done degenerate per cycle in the capacitor 

d\(\omega \) = pdt = VI dt = V₀I_o sin \(\omega \)t cos \(\omega t\)dt 

= V₀I₀ / 2 sin 2 \(\omega \)t dt

Average power degenerate per cycle in a capacitor 

Therefore, the a.c degenerate per cycle in a capacitor is zero.

Ques. 

1. A metallic rod of length l is moved perpendicular to its length with velocity v in a magnetic field B\(\to \) acting perpendicular to the plane in which the rod moves. Derive the expression for the induced EMF
2. A wheel with 15 metallic spokes each 60 cm long, is rotated at 360 rev/min in a plane normal to the horizontal component of the earth’s magnetic field. The angle of dip at that place is 60°. If the EMF induced between the rim of the wheel and the axle is 400 mV, calculate the horizontal component of the earth’s magnetic field at the place. How will the induced EMF change, if the number of spokes is increased? (All India 2016)

Ans.

1. The magnitude of the EMF, bring forth across a length dr of the rod, as it moves at right angles to the magnetic field, as suggested
   dε = Bvdr

In lieu of the actual difference across the resistor being equal to the induced EMF and equal Bx, if θ is the angle between the rod and the radius of the circle at P at time t, the surface area of the sector OPQ is acknowledged by

R = radius of the circle

2. The equation = (n) = 15,
   l = 60 cm = 60 × 10^-2 m

= If the number of spokes is increased, there would be no effect on induced EMF.

Read More: Electromagnetism

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