1. Obtain the expression for the magnetic field on the axis of a finite solenoid carrying *n *turns per unit length.

The turns of the coil are shown below with a dot indicating that the current is coming out of the page and a cross indicating that it is going into the page. Consider a point P on the axis of the solenoid.

Let *a* be the radius of the solenoid. Consider a length dx at a distance x from P. The amount of current that this section contains is . This section can be considered as a circular loop of radius *a*, the field due to which at P is given by

Let , so that . The denominator of the above expression is given by

The field due to the solenoid at P is given by

As the length of the solenoid approaches infinity both the angles and we get back as expected.

The force on the two shorter sides cancel because they carry current in the opposite direction and the force on each current element on one of the sides is canceled by the force on the corresponding element on the other side. The force of attraction on the nearer side is and the repulsive force on the far
side is . Thus the net force is attractive and is given by

3. A single current loop is in the shape of a right triangle of sides *a, b, c*. A uniform magnetic field of strength is directed parallel to the hypotenuse. Calculate the total force on the loop as well as the torque acting on the loop.

Let the triangle be in the xy plane. The unit vector along c is . Thus,

Force on side c is zero. Force on side *a* is . It can be net force is zero. This is also seen by adding up the force vectoriall since. The torque is not zero however. As the torque can

4. A semicircular loop of radius R lies in the xy plane carrying a current I in the counterclockwise direction. Magnetic field B points perpendicular to the diameter (along the y direction). Calculate the force on the straight-line segment as well as on the semi circular arc.

The force on the straight-line segment is .

on the length element is

Integrating, the force on the arc is

(Of course, the net force on the loop has to be zero!