**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 2*B*

(b) directed horizontally rightwards and has magnitude 2*B*

(c) directed vertically upwards and has magnitude 4*B*

(d) directed vertically downwards and has magnitude 4*B*

(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) μ_{0}*I*/2π*d*, directed normally into the plane of the figure, away from the reader

(b) μ_{0}*I*/4π*d*, directed normally into the plane of the figure, away from the reader

(c) μ_{0}*I*/2π*d*, directed normal to the plane of the figure, towards the reader

(d) μ_{0}*I*/4π*d*, 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) 10μ_{0}

(b) 20μ_{0}

(c) 50μ_{0}

(d) 100μ_{0}

(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) 10μ_{0}

(b) 20μ_{0}

(c) 100μ_{0}

(d) 200μ_{0}

(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) (μ_{0}*I*/2π*r*)[(*c*^{2 }–^{ }*r*^{2}) /(*c*^{2 }–^{ }*b*^{2})]

(c) (μ_{0}*I*/2π*r*)[(*c*^{2 }–^{ }*b*^{2}) /(*c*^{2 }–^{ }*r*^{2})]

(d) μ_{0}*I*/2π*r*

(e) (μ_{0}*I*/2π*r*)[(*c*^{2 }–^{ }*b*^{2}) /(*c*^{2 }–^{ }*a*^{2})]

**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) μ_{0}*I*/2π*r*

(c) (μ_{0}*I*/2π*r*)[(*c*^{2 }–^{ }*b*^{2}) /(*c*^{2 }–^{ }*r*^{2})]

(d) (μ_{0}*I*/2π*r*)[(*c*^{2 }–^{ }*r*^{2}) /(*c*^{2 }–^{ }*b*^{2})]

(e) (μ_{0}*I*/2π*r*)[(*c*^{2 }–^{ }*b*^{2}) /(*c*^{2 }–^{ }*a*^{2})]

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 4*B* [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* = (μ_{0}*I*/4π*r*) (sinΦ_{1 }– sinΦ_{2}) where Φ_{1 }and Φ_{2} are shown in the adjoining figure

The straight lines joining the point P to the ends of the conductor make the same angles (Φ_{1 }= Φ_{2} = π/2) so that sinΦ_{1 }– sinΦ_{2 }= 0. Thus* B* = 0].

The vertical portion of the conductor in the problem produces a magnetic field of magnitude μ_{0}*I*/4π*r* directed normally into the plane of the figure [Option (b)].

[*B = *(μ_{0}*I*/4π*r*) (sinΦ_{1 }– sinΦ_{2}) = (μ_{0}*I*/4π*r*) (sin π/2_{ }– sin 0) = μ_{0}*I*/4π*r*]

**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 μ_{0}*nI*/2*a* where μ_{0 }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* = μ_{0}*n*_{1}*I*_{1}/2*a*_{1} – μ_{0}*n*_{2}*I*_{2}/2*a*_{2} = (μ_{0}×20×1)/(2×0.1) – (μ_{0}×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* = μ_{0}*n*_{1}*I*/2*a*_{1} – μ_{0}*n*_{2}*I*/2*a*_{2} = (μ_{0}×20×2)/(2×0.1) – (μ_{0}×20×2)/(2×0.2) = 100μ_{0}

**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 µ_{0 }times the total current *I* passing through the surface enclosed by the closed curve. This is stated mathematically as

∫**B. d****ℓ = **µ_{0}*I * (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. d****ℓ = **µ_{0 }[*I *+ ε_{0} (dφ_{E}/dt)], where ε_{0} (dφ_{E}/dt) is the displacement current resulting from the rate of change of electric flux φ_{E}. ε_{0} 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. d****ℓ = **µ_{0}*I * 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*×2π*r* an we have

*B*×2π*r = *µ_{0}*I*

Therefore *B* = µ_{0}*I* /2π*r*

The correct option is (b).

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