AP Physics– Free Response Question (for practice) on Electromagnetic Induction

To build your confidence in answering free response questions, you may try the following practice question on electromagnetic induction:

The adjoining figure shows two smooth, straight, conducting, horizontal, parallel rails KL and MN on which a straight conducting rod PQ of mass ‘m’ can slide without friction, in a direction parallel to the rails. The rails and the rod have negligible resistance; but, there is a resistance R connected between the rails at one end. A uniform magnetic field of flux density B tesla is applied vertically downwards, throughout the region. The separation between the rails is L₁ and the length of the rod PQ is L₂.

(a) The rod PQ is pulled along the rails by a constant horizontal force F applied parallel to the rails. Derive an expression for the terminal velocity (v_terminal) of the rod.

(b) If the rod starts from rest with an initial acceleration ‘a₁ ’, find its acceleration when the velocity of the rod is v_terminal/4.

(c) What is the power dissipated in the resistance R in terms of the magnetic field and other known parameters, when the terminal velocity is attained?

(d) If the rod PQ is stationary and the magnetic field B is decreasing, will the direction of ihe induced current in the loop KPQM be clockwise or anticlockwise, when viewed along the direction of the magnetic field? Give reasons for your answer.

The above question carries 15 points which are divided among the parts (a), (b) and (c)and (d) as 6 + 5 + 2 + 2. Try to answer the above question within 15 minutes or less. I’ll be back with the answer shortly.

AP Physics B & C - Multiple Choice Practice Questions on Electromagnetic Induction

"Live as if you were to die tomorrow. Learn as if you were to live forever."

– Mahatma Gandhi

Let us discuss some multiple choice questions on electromagnetic induction.

Questions (1) and (2) are meant for AP Physics B as well as AP Physics C aspirants where as question (3) is specifically meant for AP Physics C aspirants.

(1) A search coil (a small plane coil of a few turns of insulated copper wire) of area A and negligible resistance has N turns in it. It is kept between the poles of an electromagnet so that the plane of the coil is perpendicular to the magnetic field B produced by the electromagnet. This search coil is connected in series with a resistance R to form a closed circuit. When the current in the electromagnet is reversed, the charge flowing through R is

(a) NAB/R (b) zero (c) NAB (d) 2NAB (e) 2NAB/R

The initial magnetic flux through the search coil is NAB. Since the current is reversed, the final flux is –NAB. Therefore, the change of flux (dФ) is NAB – (–NAB) = 2NAB.

Since the charge, Q = dФ/R, we obtain Q = 2NAB/R.

Suppose the current is not reversed, but instead, the search coil is rotated through 180º. In this case also the flux change will be 2NAB and the charge flowing will be 2NAB/R.

[Reversing the current through the coil of an electromagnet is not a desirable practice. The emf induced in the coil will be very large in the case of large electromagnets].

You may be asked to find the emf induced (instead of the charge) in the coil. In that case, the time in which the search coil is rotated through 180º is to be given. If the time is ‘t’ the induced emf will be 2NAB/t.

(2) Two identical parallel straight conductors P and Q are placed symmetrically on two smooth conducting rails R₁ and R₂ as shown. A uniform magnetic field exists in the region. The direction of the magnetic field is perpendicular to the plane of the figure and is inwards, as shown (by the arrow tails). If the conductor P is moved along the rails towards Q, keeping it perpendicular to the rails, then the conductor Q will

(a) remain stationary

(b) move towards P

(c) move away from P

(d) rotate in a clockwise direction

(e) rotate in an anticlockwise direction

You can easily find the correct option (c) by applying Lenz’s law. When the conductor P is moved towards Q, the magnetic flux linked with the circuit formed by the rails and the conductors P and Q is decreased since the area of the circuit is decreased. This decrement in magnetic flux is to be opposed in accordance with Lenz’s law. So, the conductor Q has to move away from P, thereby increasing the area of the circuit.

The above explanation is enough for finding the answer to the question. But it is better to see how the movement of the conductor Q happens:

Because of the flux change, an induced current flows in the circuit. The direction of the induced current through the conductor Q results in a magnetic force on Q. Since the force has to be away from P as demanded by Lenz’s law, the current in the circuit has to flow in the clockwise direction (as given by Fleming’s left hand rule). If the direction of the magnetic field is opposite, the correct option will still be (c). But, the current flowing in the circuit will be anticlockwise.

(3) A time varying magnetic field given by B(t) = 2t² – 4t +2 exists in a region where a plane coil of area 0.2 m² having 20 turns is placed with its plane perpendicular to the field. The induced emf in the coil at t = 2 s is

(a) zero (b) ) 0.8 V (c) 1.6 V (d) 16 V (e) 160 V

The induced emf is dФ/dt. Here Ф = NAB = 20×0.2(2t² – 4t +2) = 4(2t² – 4t +2)

Therefore, dФ/dt = 4(4t – 4) = 16t – 16.

Substituting for t = 2s, dФ/dt = 16 volt.

You will find more questions (with solution) on electromagnetic induction here
as well as here at physicsplus

AP Physics B&C – Equations to be Remembered in Electromagnetic Induction

Questions on electromagnetic induction are generally simple and interesting. You have to remember the following things in this section for answering the questions (especially multiple choice questions) within the stipulated time:

1. Magnetic flux Ф (in weber) through a plane coil of area A (in m²) and number of turns N is given by Ф = NABcosθ where B is the magnetic flux density (in tesla or weber/m²) and θ is the angle between the vectors A and B. Note that the area vector is perpendicular to the plane of the area (fig).

If B is perpendicular to the plane of the area, the angle θ is zero and the magnetic flux is maximum, equal to NAB. If B is parallel to the plane of the area, the angle θ is 90º and the flux is zero.

2. Induced emf V(in volt) as given by Faraday’s law: V = – dФ/dt where dФ is the change in magnetic flux in a time dt.

The negative sign is because of Lenz’s law which says that the induced emf always opposes the change which produces it.

3. Induced charge (Q coulomb) flowing in a circuit: Q = Change of magnetic flux (in weber)/ Resistance (in ohm) of the circuit.

Therefore, Q = dФ/R

Note that the induced charge does not depend on the time in which the flux change occurred.

4. Motional emf (V) induced in a straight conductor of length L moving with velocity ‘v’ perpendicular to a magnetic field of flux density B is given by

V = BLv

5. Motional emf (voltage) induced between the ends of a straight rod of length L rotating about a perpendicular axis through one end in a magnetic field B acting perpendicular to the plane of rotation is given by

V = ½ BL²ω

where ω is the angular velocity of the rod.

If a metal disc of radius R rotates (about a central axis) with its plane perpendicular to a magnetic field B, the induced voltage between the centre and the edge of the disc is obtained by replacing L in the above expression with R so that

V =½ BR²ω

Note that the motional voltage ( 4 & 5 above) induced is numerically equal to the product of B and the area swept per second.

6. The emf induced in a plane coil of area A and number of turns N rotating with angular velocity ω in a magnetic field B acting perpendicular to the axis of rotation is given by

V = NABω sin ωt

(if the time starts when the plane of the coil is perpendicular to B).

This is an alternating emf.

If the time is reckoned from the instant the plane of the coil is parallel to the magnetic field B, the above expression gets modified as

V = NABω cos ωt

7. Ideal transformer equations:

V₂/V₁ = N₂/N₁ = I₁/I₂ where V₁ and V₂ are the primary and secondary voltages, I₁ and I₂ are the primary and secondary currents and N₁ and N₂ are the primary and secondary turns respectively.

8. Efficiency (η) of a transformer:

η = V₂I₂/ V₁I₁

Percentage efficiency is (V₂I₂/ V₁I₁) ×100.

Questions on electromagnetic induction will be discussed in the next post, following which equations as well as questions involving inductance will be discussed.

Fluid Mechanics- Questions on Bernoulli’s Theorem

As promised in the post dated 4^th December 2007, I give below more questions from Fluid Mechanics. Consider the following question:

In the figure, a tall cylindrical empty tank with a small side hole of square shape near its bottom is shown. The area of the hole is A. Water flowing with a velocity ‘v’ through a pipe of circular cross section having the same area A (as that of the hole) starts falling vertically down in to the tank. Which one of the following statements is correct?

(a) No water can remain in the tank

(b) The height of water column in the tank will go on increasing

(c) The height of water column in the tank will increase initially and will remain steady at v/2g

(d) The height of water column in the tank will increase initially and will remain steady at √(v/2g)

(e) The height of water column in the tank will increase initially and will remain steady at v²/2g.

The velocity with which water will flow out through the side hole in the tank will be small initially and hence the height of water column will go on increasing until the volume of water flowing into the tank becomes exactly equal to the volume flowing out through the side hole. The height of water column remains fixed at the value ‘h’ for attaining this condition. Since the area of the hole is equal to the area of cross section of the inlet pipe, the steady height condition is attained when the velocity of efflux (of water through the hole), as given by Torricelli’s theorem, is equal to the velocity of water through the inlet pipe.

Therefore we have, v = √(2gh) from which h = v²/2g [ Option (e)].

Now, consider the following MCQ:

Water is flowing steadily out through the end of a vertical pipe (fig) with a velocity of 4 ms^–1. At what distance from the end of the pipe will the area of cross section of the stream of water be (2/3)A where A is the area of cross section of the pipe?

(a) 0.5 m (b) 1 m (c) 1.5 m (d) 2.2 m (e) 2.5 m

From the equation of continuity, we have A₁v₁ = A₂v₂.

Here A₁ = A, A₂ = 2A/3 and v₁ = 4 ms^–1 so that v₂ = 6 ms^–1. This is the velocity of the stream at distance ‘x’ (let us say) below the end of the pipe, where the area of cross section of the stream reduces to (2/3)A

Since the stream is open to the atmosphere, the pressures are the same at all regions of the stream. The kinetic energy of the water particles will be increased because of an equal decrease in the gravitational potential energy.

Considering a mass m of water, we can write ½ mv₂² = ½ mv₁² + mgx. [Note that we have taken the reference (zero) level for the gravitational potential energy at distance x below the end of the pipe].

This gives

v₂ = √(v₁² + 2gx). Or, 6 = √(4² + 2gx).

Taking the acceleration due to gravity to be 10 ms^–2, x = 1m.

[If you blindly write Bernoulli’s equation, you will have P + ρgx + (½)ρv₁ ² = P + (ρg×0) + (½)ρv₂² from which you will get v₂ = √(v₁² + 2gx). You will then proceed to find ‘x’].

Fluid Mechanics- Multiple Choice Practice Questions on Force of Buoyancy

In the post dated 3^rd December 2007, the essential formulas you have to remember in the section 'Fluid Mechanics' were discussed. As promised, I will give you a few multiple choice questions with solution. Here is a question on force of buoyancy:

In the adjoining figure, a metallic block of volume 20 cm³ is shown suspended, using a thin nylon string, from an independent support. The entire block is immersed in water in a beaker, without touching the sides of the beaker. The reading of the electronic balance (on which the beaker is placed) is 257 gram, as shown.

If the weight of the beaker is 37 gram, what is the weight of the water in the beaker?

(a) 200 g (b) 220 g (c) 230 g (d) 240 g (e) Data insufficient for calculation

Since the metallic block is completely immersed in water, the force of buoyancy on the block is equal to the weight of 20 cm³ of water displaced by the block. In other words, water exerts an upthrust equal to the weight of 20 g of water. The block exerts an equal force downwards (reaction force) which the electronic balance registers in addition to the weight of the beaker and water.

The weight of the beaker and water together is therefore 257–20 = 237 g. Since the weight of the beaker is 37 g, the weight of water alone is 237–37 = 200 g.

Now, suppose the block was kept immersed using a rigid support instead of the string. Then also, the same explanation holds good. If you dip your finger gently in the water in the beaker, the reading of the balance will increase since the water exerts a force of buoyancy on your finger and your finger exerts an equal and opposite force of reaction.

Here is another multiple choice question (MCQ):

An iceberg floating in sea water just sinks when a mass of 360 kg is placed on it. What is the mass of the iceberg? (Relative density of sea water = 1.02; relative density of ice = 0.9)

(a) 1500 kg (b) 1800 kg (c) 2400 kg (d) 2700 kg (e) 3200 kg

The density of ice is 900 kg m^–3 and that of sea water is 1020 kg m^–3, as obtained from their relative densities. When the iceberg just sinks, the mass of displaced sea water is 1200V kg. where V is the volume of the iceberg. The mass of iceberg is 900V kg. Therefore we have, in this case (since the weight of floating body = weight of displaced liquid),

900V + 360 = 1020V, from which V = 3 m³

Therefore, mass of iceberg = 900×3 = 2700 kg.

Let us consider one more question:

A solid sphere of volume V remaining submerged in a liquid of density σ is lifted through a height ‘x’ within the liquid. If the density of the material of the sphere is ρ where ρ > σ, which one of the following statements is correct?

(a) There is no change in the potential energy of the sphere

(b) The potential energy of the sphere decreases by Vρgx

(c) The potential energy of the sphere increases by Vρgx

(d) The potential energy of the sphere increases by V[(ρ+σ)/2]gx

(e) The potential energy of the sphere increases by V(ρ – σ)gx

You will have to exert a force equal to the apparent weight of the sphere for lifting it. The apparent weight of the sphere is Vρg – Vσg = V(ρ – σ)g. The first term is its real weight and the second term is the up thrust (force of buoyancy) on it.

The work done in lifting the sphere through the height ‘x’ is V(ρ – σ)gx and this is the increase in potential energy of the sphere.

More questions from Fluid Mechanics will be discussed in the next post.

Meanwhile see a useful post in this section here.