Electromotive Force: Important Questions

Electromotive Force Important Questions are provided in the article. EMF stands for electromotive force. EMF is the voltage at the terminals of the source in the absence of an electric current.

• Electromotive force is the transfer of energy per unit charge in an electrical circuit.
• It is measured in volts.
• A device named a transducer provides emf by converting other forms of energy into electrical energy.
• Batteries also produced emf.
• Electromotive force is not a physical force.

Electromotive Force

The video below explains this:

Electromotive Force Detailed Video Explanation:

Very Short Answer Questions (1 Mark Question)

Ques. The electromotive force of a cell is basically work. Explain how.

Ans. EMF= work done per unit charge = Fd/q ​= qEd/q ​= Ed = V

Ques. An electric generator converts mechanical energy into which form of energy?

Ans. An electric generator converts mechanical energy into electrical energy.

Ques. What is an electrochemical cell?

Ans. An electrochemical cell is a device used to produce electric energy from chemical reactions.

Ques. A flexible wire of irregular shape, abed as shown in the figure, turns into a circular shape when placed in a region field which is directed normally to the plane of the loop away from the reader. Predict the direction of the induced current in the wire. [Foreign 2014]

Ans. The wire is expanding to form a circle, which means that force is acting outwards on each part of the wire because of the magnetic field (acting in the downward direction). The direction of the induced current should be such that it will produce a magnetic field in the upward direction (towards the reader). Hence, the force on the wire will be in an inward direction, i.e. induced current is flowing in an anti-clockwise direction in the loop.

Ques. A pair of adjacent coils have a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20A in 0.5 secs, what is the change of flux linkage with the other coil?

Ans. Given

• Mutual inductance, M = 1.5 H
• Change in current, di = 20 – 0 = 20 A
• Time, dt = 0.5 s

We have induced emf, e = dϕ/ dt

Where dϕ is the change in the flux linkage with the coil

Emf is related to mutual inductance as e= Mdi/ dt  

Equating both equations, we get

dϕ/ dt = Mdi/ dt

⇒ dϕ = Mdi

⇒ dϕ = 1.5 x 20

⇒ dϕ = 30 Wb

Ques. What is the formula of electromotive force?

Ans. The formula of emf is

ε= V + Ir

Where

• ε = electromotive force 
• V = voltage of the cell
• I = current across the circuit 
• r = Internal resistance of the cell.

Ques. Consider a magnet surrounded by a wire with an on/off switch S (figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain. (NCERT Exemplar)

Ans. There is no relative motion between the magnet and the coil. This means that there is no change in magnetic flux, hence no electromotive force is produced and hence no current will flow in the circuit.

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Short Answer Questions (2 Marks Question)

Ques. Electromotive force represents energy per unit charge. Explain.

Ans. Electromotive force is the potential difference (or voltage V) of the cell.

Using V=qW​

where W is the work done in moving a charge q from one point to another having potential difference V.

Thus electromotive force represents energy per unit charge.

Ques. A load of 20 ohms is connected with a battery of EMF, ε = 6 volts having the internal resistance of the battery being 4 ohms. Calculate the terminal potential difference.

Ans. Given

• Emf of the battery, ε = 6 volts
• External load, R = 20 ohm
• The internal resistance of the battery, r = 4 ohm

According to Ohm’s law, the current flowing through the circuit is given by

I = ε/R+r

⇒ I = 6/(20+4) = 6/24 = 0.25 A

Now, the terminal potential difference is given by

V = ε – Ir

⇒ V = 6 – (0.25 x 4)

⇒ V = 5 volt

Ques. State the law that gives the polarity of the induced emf. [All India 2009]

Ans. Lenz’s law gives the polarity of induced emf. The induced emf (or current) always opposes the cause that produces it. Lenz’s law The induced emf of the induced current is a circuit that always opposes the cause that produces it.

When the North pole of the bar magnet approaches the loop, the induced current in the coil is anti-clockwise (forming the North pole of the current loop) when viewed from the magnetic side. The face of the coil being the North pole opposes the arrival of the North pole of a magnet. Hence, opposes the cause that produces it. Also, a certain amount of work has to be done by an external agency to bring the North pole near the coil against of force of repulsion applied by the induced current loop. The work done by external agencies appears in the form of electrical energy. So, Lenz’s law is a consequence of the principle of the law of conservation of energy.

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Long Answer Questions (3 Marks Question)

Ques. A rod of length / is moved horizontally with a uniform velocity v in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod. [All India 2014]

Ans. Consider a straight conductor moving with velocity v and a U-shaped conductor placed in the perpendicular magnetic field as shown in the figure.

Ques. (i) Derive an expression for the drift velocity of free electrons.

(ii) How does the drift velocity of electrons in a metallic conductor vary with an increase in temperature? Explain. (All India)

Ans. (i) Expression for drift velocity: When a potential difference is applied across a conductor, an electric field is produced, and free electrons are acted upon by an electric force (= -Ee). Due to this, electrons accelerate and keep colliding with each other and acquire a constant (average) velocity v_d

(ii) The drift velocity of electrons decreases with temperature because the time of relaxation decreases.

Ques. A solenoid with 15 turns/cm has a small loop of area 2 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid charges steadily from 2 A to 4 A in 0.1 sec, what is the induced emf in the loop while the current is charging?

Ans. The no.of turns = 1500 turns/m

small loop area of solenoid A = 2 cm²

Change in current = di = 4 -2 = 2A

= dt = 0.1 s

Induced emf as per Faraday’s law

= e= dϕ/ dt  — (1)

where ϕ = induced flux through the small loop =BA — (2)

= B = μ₀ni — (3)

= μ₀ = 4π x 10^-7 H/m

Reducing (1)

= e = d/dt (BA)

= e = Aμ₀n x (di/dt)

= e = 2 x 10-2 x 4π x 10^-7 x 1500 x (2/0.1)

= e = 7.54 x 10^-6 V

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