Question

a long solenoid(1000 turns/m) carries a current of 25 mA and has an inside radius of 2.0 cm. A long wire which is parallel to and 4.0 cm from the axis of the solenoid carries a current of 6.0 A. Determine the magnitude of the magnetic field at a point on the axis ofthe solenoid.

Answer 1

First, calculate the field on the axis of a solenoid (blue):

$$B_s=\mu_0nI_s.$$

Also, calculate the field produced by the wire (red) with a current at a distance d=6 cm from the wire:

$$B_w=\frac{\mu_0I_w}{2\pi d}.$$

Draw the figure.

The two vectors (from the field and the wire) are perpendicular to each other. The resulting (green) vector magnitude will be

$$B=\sqrt{B_s^2+B_w^2}=\mu_0\sqrt{(nI_s)^2+\frac{I_w^2}{(2\pi d)^2}}=\\=3.7\cdot10^{-5}\text{ T}.$$