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 C – Applications of Ampere’s Circuital Law – Magnetic Field due to Straight Infinitely Long Thick Current Carrying Cylinders and Pipes

In the post dated 1^st February 2008 the equations to be noted in connection with magnetic fields were given. [To access that post click here]. Ampere’s circuital law was briefly mentioned there.

Ampere’s circuital law 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].

In this post I want to make you realize the importance of Ampere’s circuital law as a simple tool for evaluating the magnetic field in situations exhibiting symmetry in such a manner that the magnitude of the magnetic field is the same along the path chosen for the line integral.

You might have used Ampere’s circuital law to derive the expression, B = µ₀µ_rnI for the magnitude of the magnetic flux density inside very long solenoids and toroids. [Here µ₀ is the permeability of free space, µ_r is the relative permeability of the medium filling the interior of the solenoid (or toroid), n is the number of turns per metre and I is the current].

You might have certainly used Ampere’s circuital law to derive the expression, B = μ₀I/2πr for the magnitude of the magnetic flux density at a point P distant ‘r’ from an infinitely long straight conductor carrying a steady current I. The above expression shows that the magnetic field is inversely proportional to the distance of the point P from the conductor. Normally you won’t worry about the variation of the magnetic field inside the conductor since the conductor is assumed to be thin. Now, suppose the conductor is thick. Now you have enough space to move about inside the body of the conductor to explore the variation of the magnetic field!

Let us use Ampere’s circuital law for finding the magnetic field due to a thick cylindrical conductor of radius R. A cross section of the thick conductor is shown in the following figure.

The conductor is straight and infinitely long, with its length normal to the plane of the figure. The current I is supposed to flow into the plane of the figure, away from the reader.

Consider a point P at distance r from the centre of the conductor such that r < R so that the point P is within the conductor. This is a situation exhibiting symmetry and the magnitude B of the magnetic flux density at all points at the same distance r from the centre must be the same. Applying Ampere’s circuital law, we equate the line integral of the magnetic field vector B over a circular path (red dotted circle) of radius r to the product of µ₀ and the appropriate current:

∫B. dℓ = µ₀I₁ ………….(i)

where I₁ is the current passing (piercing) through the surface enclosed by the circular path of radius r. Note that I₁ is the current carried by the cylindrical portion of the conductor having radius ‘r’so that

I₁ = I(πr²/πR²) = Ir²/R²

[The total current I passes through the total cross section area πR². So the current I₁ passing through the cross section area πr² is I(πr²/πR²) = Ir²/R²].

Substituting for I₁ in equation (i) and remembering that the direction of the magnetic field B is along the path of integration everywhere, we have

2πrB = µ₀Ir²/R²

Therefore, B = μ₀Ir/2πR²

This shows that the magnetic field at the centre of the conductor is zero (since r = 0) and it increases linearly with the increase in distance r within the conductor. The maximum value of field is at the surface of the conductor (corresponding to r = R) and is equal to μ₀I/2πR.

The direction of the field lines is clockwise as we have considered the current to flow into the plane of the figure, away from the reader.

The magnitude of magnetic flux density at a point such as P’ outside the conductor (at distance r > R) is given by

2πrB = µ₀I since the entire current I passes (pierces) through the surface enclosed by the circular path of radius r. This gives

B = μ₀I/2πr

[This is the usual expression for the magnetic field due to a long straight (thin) current carrying conductor].

The variation of the magnitude of the magnetic flux density B against the distance r is shown graphically in the adjoining figure.

Now we will consider the case of a long, straight, thick cylindrical current carrying pipe of inner radius R₁ and outer radius R₂.

[Derivation of the magnetic field due to a current carrying pipe using Ampere’s circuital law was one of the free response questions in the AP Physics C 2011 question paper

A cross section of the pipe is shown in the adjoining figure. The current I is supposed to flow into the plane of the figure, away from the reader.

Consider a point such as P at distance r such that R₁ < r < R₂. To find out the magnetic flux density B at the point P within the material of the pipe, we consider the line integral of B over a circular path (red dotted circle) of radius r and apply Ampere’s circuital law to obtain

∫B. dℓ = µ₀I₁ ………….(i)

where I₁ is the current passing (piercing) through the surface enclosed by the circular path of radius r.

We have I₁ = Iπ(r² – R₁²) /π(R₂² – R₁²) = I(r² – R₁²) /(R₂² – R₁²)

[The total current I passes through the total cross section area π(R₂² – R₁²). So the current I₁ passing through the cross section area π(r² – R₁²) is Iπ(r² – R₁²) /π(R₂² – R₁²) = I(r² – R₁²) /(R₂² – R₁²)].

Substituting for I₁ in equation (i) and remembering that the direction of the magnetic field B is along the path of integration everywhere, we have

2πrB = µ₀I(r² – R₁²) /(R₂² – R₁²). This gives

B = µ₀I(r² – R₁²) /2πr(R₂² – R₁²)

Or, B = (µ₀I/2πr) [(r² – R₁²)/(R₂² – R₁²)]

This expression shows that the magnetic field is zero at r = R₁ (at the inner surface of the pipe) and it increases non-linearly (unlike in the case of a solid cylindrical conductor), within the material of the pipe.

Now, what is the magnetic field inside the hollow region of the pipe? You might have definitely come across this question earlier and might have proved (by other methods) that it is zero everywhere within the hollow region. Well, Ampere’s circuital law offers an immediate proof to you:

When you consider the line integral over a concentric circular path (of the type we considered above) within the hollow region, you find that there is no current passing (piercing) through the surface enclosed by the circular path. Therefore we have

∫B. dℓ = µ₀×0 so that B = 0

What is the magnetic field at a point outside the pipe?

We consider a point such as P’ at distance r > R₂. The magnitude of magnetic flux density B at the point P’ is given by

2πrB = µ₀I since the entire current I passes (pierces) through the surface enclosed by the circular path of radius r. This gives

B = μ₀I/2πr

[As you might have expected, this is same as the usual expression for the magnetic field due to a long straight (thin) current carrying conductor].

The nature of variation of the magnitude of the magnetic field B with the distance r is shown in the adjoining figure.

You will find an interesting question with solution in this section here.