Question

a) Describe how a circular turn (or a circular coil) constitutes a dipole magnet.

b) Consider a solenoid of N turns with a radius r and length l, and carrying a current I. What is the magnitude of magnetic field inside the solenoid (but not near the ends)? Does this magnitude depend on distance the center of the solenoid? ?

c) What is the magnitude of magnetic field outside the solenoid (but not near the ends)?

Answer 1

1) Consider a circular coil of radius a carrying a steady current I. Let it be placed with its center at origin and plane perpendicular to the X axis.

Now imagine a small element AB of length dl. The magnetic field at a point P on the X axis at a distance x from the origin is given by

where r is the distance between the element and the point P.

If vector r makes an angle alpha with the X axis, then dB will be inclined to the +Y axis. Hence dB can be resolved into X and Y components given by

Now if we consider an identical element CD diametrically opposite to AB, the magnetic field due to CD will be in a direction inclined to the Y axis.

The above expressions makes it clear that the Y component of magnetic field due to diametrically opposite identical elements cancels each other. Hence the net magnetic field at P is simply the vector sum of x components of all the elements of the ring.

Thus the magnetic field due to a coil at a point on the x axis depends on the current I, the radius of the coil and the distance of the point from the center.

If the point P selected at the center of the coil then x=0

The direction of magnetic field due to a circular coil is obtained by the right hand thumb rule which is stated as follows:

Curl the palm of right hand around the circular coil with the fingers pointing in the direction of current. Then the extended thumb gives the direction of magnetic field.

Thus an anti-clock wise current gives a magnetic field out of the coil and a clock wise current gives a magnetic field into the coil.

Current loop as a magnetic dipole

In the preceeding section we have seen that anti clock wise current gives outgoing flux, indicating south polarity. In short a current loop gives north polarity on one side and south polarity on the other side resulting in a magnetic dipole.

The magnetic dipole moment of a current loop is defined as the product of electric current and area.Its SI unit is Am2.

Now magnetic field at a point on the axis of a circular coil given by the equation 10 can be rewritten as

The above equation is similar to that of the electric field on the axial line of an electric dipole. This we can conclude that a current loop is equivalent to a magnetic dipole of moment equal to m=IA. This can be generalized to any geometric shape of the current loop.

2) A solenoid is made out of a current-carrying wire which is coiled into a series of turns (with the turns preferably as close together as possible). The magnetic field due to a straight length of wire is shown in Figure 1 - the field circles the wire and its magnitude (or strength) decreases with radial distance from the wire.

Figure 1: Magnetic field due to a straight wire

In a solenoid, a large field is produced parallel to the axis of the solenoid (in the z-direction in figure 2). Components of the magnetic field in other directions are cancelled by opposing fields from neighbouring coils. Outside the solenoid the field is also very weak due to this cancellation effect and for a solenoid which is long in comparison to its diameter, the field is very close to zero. Inside the solenoid the fields from individual coils add together to form a very strong field along the center of the solenoid.

Figure 2: Magnetic field in a solenoid

To calculate the magnitude of the field in the solenoid, we used Ampere's law. Ampere's law relates the circulation of B around a closed loop to the current flux through the loop x