Sol. (i) If a large plate carries a current which is uniformly distributed over its width, then a uniform magnetic field is established around it.

If a section of plate, which is normal to the direction of flow of current, is considered then it will be as shown in fig.

Let magnetic induction of the field induced due to current in one plate be B.

Considering a length l in the section as shown in fig. and applying Amperes's Circuital law,

B. 2l = u 0 (lj)

or B = µ 0 j

But there are two plates which carry equal current but in opposite directions. Therefore, magnetic fields due to these currents, in the space between the plates are unidirectional.

Resultant magnetic field induction between the plates = 2B = µ 0 j

Now consider an elemental width dx in the section of upper plate as shown in fig. This elemental width is similar to a long straight conductor carrying current di = j dx

Magnetic induction at this conductor due to current in lower plate is B = µ 0 j (leftward)

Hence, force on this conductor, dF = B di per unit length

or dF = µ 0 j 2 dx

per unit length

But area of unit length of the conductor considered = 1. dx = dx

Force per unit area of upper plate =

= µ 0 j 2