AP Physics B & C - Multiple Choice Practice Questions on Magnetic Fields due to Current Carrying Conductors

A few multiple choice questions (for practice) related to magnetic fields produced by current carrying wires are given below. You may solve them yourself and check your answers by referring to the solution given below the set of questions.

Question No.1:

A plane square loop of wire (Fig.) carrying a current is oriented with its plane horizontal. On viewing from above, the current in the loop flows in clockwise direction. If the magnitude of the magnetic flux density at the centre of the loop due to each side is B, the resultant magnetic flux density at the centre of the loop is

(a) directed horizontally leftwards and has magnitude 2B

(b) directed horizontally rightwards and has magnitude 2B

(c) directed vertically upwards and has magnitude 4B

(d) directed vertically downwards and has magnitude 4B

(e) zero

Question No.2:

A straight infinitely long wire carrying a current I is given a 90º bend at the position O (Fig.) near its middle. What is the magnetic flux density at the point P (shown in the figure) at distance d from the bend?

(a) μ₀I/2πd, directed normally into the plane of the figure, away from the reader

(b) μ₀I/4πd, directed normally into the plane of the figure, away from the reader

(c) μ₀I/2πd, directed normal to the plane of the figure, towards the reader

(d) μ₀I/4πd, directed normal to the plane of the figure, towards the reader

(e) Zero

Question No.3:

Two coplanar concentric circular coils P and Q of 20 turns each carry currents of 1A and 2 A respectively in opposite directions. If their radii are 10 cm and 20 cm respectively, what is the magnitude of the resultant magnetic flux density at their common centre?

(a) 10μ₀

(b) 20μ₀

(c) 50μ₀

(d) 100μ₀

(e) Zero

Question No.4:

If the coils in question no.3 carry the same current of 2A (in opposite directions), what will be the magnitude of the resultant magnetic flux density at their common centre?

(a) 10μ₀

(b) 20μ₀

(c) 100μ₀

(d) 200μ₀

(e) Zero

The above questions are meant for AP Physics B as well as AP Physics C.

The following questions (5 and 6) are meant specifically for AP Physics C.

Question No.5:

An infinitely long coaxial cable has an inner central cylindrical conductor of radius a and an outer conducting cylindrical pipe of inner radius b and outer radius c (A portion of the coaxial cable is shown in the adjoining figure). It carries equal and opposite currents of magnitude I on the inner an outer conductors. What is the magnitude of the magnetic flux density at a point P outsie the coaxial cable at distance r from the axis?

(a) Zero

(b) (μ₀I/2πr)[(c² – r²) /(c² – b²)]

(c) (μ₀I/2πr)[(c² – b²) /(c² – r²)]

(d) μ₀I/2πr

(e) (μ₀I/2πr)[(c² – b²) /(c² – a²)]

Question No.6:

In the case of the coaxial cable of question no.5 above, what is the magnitude of the magnetic flux density at a point P in between the central conductor and the outer pipe, if the distance of the point P from the axis is r?

(a) Zero

(b) μ₀I/2πr

(c) (μ₀I/2πr)[(c² – b²) /(c² – r²)]

(d) (μ₀I/2πr)[(c² – r²) /(c² – b²)]

(e) (μ₀I/2πr)[(c² – b²) /(c² – a²)]

Answers to the above questions are given below:

Answer to Question No.1:

The magnetic fields due to all the four sides of the loop act vertically downwards at the centre of the loop and they add up to produce a resultant field of magnitude 4B [Option (d)].

Answer to Question No.2:

The magnetic field at P due to the horizontal portion of the conductor is zero since the point P is lying on the straight line indicating the direction of flow of the current.

[The magnitude B of the magnetic field at a point P due to a finite length of straight conductor is generally given by

B = (μ₀I/4πr) (sinΦ₁ – sinΦ₂) where Φ₁ and Φ₂ are shown in the adjoining figure

The straight lines joining the point P to the ends of the conductor make the same angles (Φ₁ = Φ₂ = π/2) so that sinΦ₁ – sinΦ₂ = 0. Thus B = 0].

The vertical portion of the conductor in the problem produces a magnetic field of magnitude μ₀I/4πr directed normally into the plane of the figure [Option (b)].

[B = (μ₀I/4πr) (sinΦ₁ – sinΦ₂) = (μ₀I/4πr) (sin π/2 – sin 0) = μ₀I/4πr]

Answer to Question No.3:

The magnetic flux ensity at the centre of a circular current carrying coil is directed along the axis and has magnitude μ₀nI/2a where μ₀ is the permeabitity of free space, n is the number of turns in the coil and a is the radius of the coil.

Since the currents in the coils are in opposite directions, the magnetic fields are in opposite directions and the magnitude B of the resultant magnetic field at the common centre of the coils is given by

B = μ₀n₁I₁/2a₁ – μ₀n₂I₂/2a₂ = (μ₀×20×1)/(2×0.1) – (μ₀×20×2)/(2×0.2) = 0

Answer to Question No.4:

The resultant magnetic field at the common centre of the coils is given by

B = μ₀n₁I/2a₁ – μ₀n₂I/2a₂ = (μ₀×20×2)/(2×0.1) – (μ₀×20×2)/(2×0.2) = 100μ₀

Answer to Question No.5:

This question can be worked out easily using Ampere’s circuital law which states that the line integral of magnetic flux density B over any closed curve is equal to µ₀ times the total current I passing through the surface enclosed by the closed curve. This is stated mathematically as

∫B. dℓ = µ₀I (The integration is over the closed path)

[Ampere’s circuital law as modified by Maxwell to accommodate the displacement current flowing through dielectrics and free space is

∫B. dℓ = µ₀ [I + ε₀ (dφ_E/dt)], where ε₀ (dφ_E/dt) is the displacement current resulting from the rate of change of electric flux φ_E. ε₀ is the permittivity of free space].

We imagine a circle of radius r, with its centre at the axis of the coaxial cable, as the closed curve for the integration. Since this circular path encloses two equal currents in opposite directions, the total current I passing through the surface enclosed by the closed curve is zero. Therefore the magnitude B of the magnetic flux density at a point P outsie the coaxial cable must be zero.

Answer to Question No.6:

In orer to find the magnetic flux density at a point P in between the central conductor and the outer pipe, we imagine a circle of radius r, with its centre at the axis of the coaxial cable. Since this circular path encloses the entire current I passing through the central conductor, we have (from Ampere’s circuital law)

∫B. dℓ = µ₀I where B is the magnetic flux density at point P at distance r . The direction of the magnetic field coincides with the circular path of integration since the magnetic field lines due to a straight conductor are in the form of concentric circles. The line integral on the left hand side of the above equation therefore simplifies to B×2πr an we have

B×2πr = µ₀I

Therefore B = µ₀I /2πr

The correct option is (b).

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

“Maturity is often more absurd than youth and very frequently is more unjust to youth.”

– Thomas A. Edison

Michael Faraday’s discovery of electromagnetic induction was a turning point in the history of mankind. When he made the first public announcement that the relative motion between a magnet and a coil of wire could cause the flow of a feeble electric current through the coil, he had to face this question: “But what is the use?” Faraday countered this with another question: “What is the use of a new born baby?”

The baby has grown rapidly to become a very healthy youth who will remain so for many more decades!

The phenomenon responsible for the generation of electric power for feeding the modern world still continues to be electromagnetic induction.

Questions on electromagnetic induction are generally interesting. Click on the label ‘electromagnetic induction’ below this post; you will find all posts on electromagnetic induction published so far on this site.

Today we will discuss a few more multiple choice practice questions in this section.

(1) Earth’s resultant magnetic field at California has magnitude B tesla and it makes an angle θ with the horizontal. Assuming that there are no other magnetic fields, what will be the voltage induced between the tips of the wings of an airplane of wing-span L flying horizontally with speed v?

(a) BLv

(b) BLv sin θ

(c) BLv cos θ

(d) BLv/sin θ

(e) BLv/cos θ

Since the airplane is flying horizontally it can ‘cut’ the vertical magnetic field lines to generate a motional emf V given by

V = B_verticalLv where B_vertical is the vertical component of earth’s magnetic field at the place.

With reference to the adjoining figure we have

B_vertical = B sin θ

Therefore, the voltage induced between the tips of the wings of the airplane is BLv sin θ.

(2) A plane square loop of thin copper wire has 100 turns. Each side of the loop is 10 cm long and the loop is oriented with its plane making an angle of 30º with a uniform magnetic field of flux density 0.4 tesla. If the loop is rotated in 0.5 second so as to orient its plane at right angles to the magnetic field, what will be the magnitude of the average emf induced in the loop?

(a) 0.1 volt

(b) 0.2 volt

(c) 0.4 volt

(d) 0.8 volt

(e) 2 volt

The induced emf V is given by

V = – dФ/dt where dФ is the change in the total magnetic flux linked with the coil and dt is the time taken for the flux change.

[The negative sign is the consequence of Lenz’s law by which the induced emf has to oppose the change of flux dФ].

Since we are required just to find the magnitude of the induced voltage, we may ignore the negative sign

Since the coil has N (=100) turns, the total flux linked with the coil is Nφ where φ is the flux per turn given by

φ = BAcos θ where B = 0.4 tesla and A = area of the square loop = (0.1)² m² = 0.01 m²

The angle θ is the angle between the magnetic field and the area vector.

[Remember that the area vector is directed perpendicular to the plane of the coil].

Since the plane of the coil makes an angle of 30º with a magnetic field, the area vector makes an angle of 60º with the magnetic field.

The initial magnetic flux linkage is NBAcos 60º = 100×0.4×0.01×(1/2) = 0.2 weber.

Since the area vector and the magnetic field are finally parallel (or anti-parallel), the final flux linkage is NBAcos 0º = 100×0.4×0.01 = 0.4

The change of flux dФ = 0.4 – 0.2 = 0.2

Therefore, induced emf = (Change of flux) /(Time) = 0.2/0.5 = 0.4 volt.

(3) Suppose that the resistance (R) of the loop in the above question is 10 Ω. What will be the induced current in the loop if the loop is kept stationary and the magnetic field is steadily reduced to zero in a time of 40 millisecond?

(a) 0.2 A

(b) 0.5 A

(c) 1 A

(d) 1.5 A

(e) 2 A

The initial magnetic flux linked with the loop (as shown above) is NBAcos 60º = 100×0.4×0.01×(1/2) = 0.2 weber.

When the magnetic field is reduced to zero, the magnetic flux is reduced to zero. Therefore the change of magnetic flux is 0.2 weber. The emf V induced in the loop is given by

V = (Change of flux) /(Time) = 0.2/(40×10^–3) volt = 5 volt.

The current induced in the loop is V/R = 5/10 A = 0.5 A [Option (b)]

The following question is meant specifically for AP Physics C aspirants:

(4) A straight conductor of length L and mass M can slide down along a pair of long, smooth, conducting vertical rails P and Q of negligible resistance (Fig.). A resistor of resistance R is connected between the ends of the rails as shown in the figure. A uniform magnetic field of flux density B acts perpendicularly into to the plane containing the rails and the sliding conductor. The terminal velocity of fall of the rod is

(a) MgR/LB

(b) mgL/B²R²

(c) B²L²/mgR

(d) mgB/L²R

(e) mgR/B²L²

When the rod slides down under gravity, the magnetic flux linked with the closed circuit comprising the rod, rails and the resistor R changes and a current is induced in the circuit. The induced emf is the motional emf BLv where v is the velocity of the rod. The induced current I in the circuit is BLv/R.

By Lenz’s law the induced current has to oppose the motion of the rod. It is the magnetic force ILB which brings in this opposition. When the velocity of the rod increases, the opposing magnetic force also increases. When the magnitudes of the gravitational force (weight Mg of the rod) and the opposing magnetic force become equal, the rod moves with a constant (terminal) velocity.

Therefore, we have

ILB = Mg

Substituting for I we have (BLv/R)LB = Mg

Or, B²L²v/R = Mg

This gives v = MgR/B²L²

Now, let me ask you a question:

If the direction of the magnetic field in the above question is reversed, will the rod still attain a terminal velocity? Think of it and arrive at the answer ‘YES’.

You will find a few more questions (with solution) in this section here.