Question

A straight, 2.5-m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth’s magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

Short answer

(a) The magnitude and direction of the force the magnetic field exert on the wire if the orientation of the current is from west to east is $2.1\times {10}^{-4}N$upward.

(b) The magnitude and direction if the current is directed from north to south is$2.1\times {10}^{-4}$ from east to west.

(c) The magnitude of the force if the current is from the north to south is 0.

(d) The magnetic field is very small hence not large enough to cause any significant effects.

Step-by-step solution

Force if the current is running from west to east

Given, a straight wire of length l = 2.5 m , carrying current of l = 1.5 A , the magnetic field is B = 0.55gauss , from south to north.

For the current in the wire running from west to east and the angle between the magnetic field and the current is$\mathrm{\Phi }=90°$

Hence the magnitude of the force is given by:

$\begin{array}{r}\mathrm{F}=\mathrm{ILBsin}\mathrm{\theta }\\ =\mathrm{IIBsin}\left({90}^{\circ }\right)\\ =\mathrm{IIB}\\ =(1.5\mathrm{A})(2.5\mathrm{m})\left(0.55\times {10}^{-4}\right)\\ =2.1\times {10}^{-4}\mathrm{N}\end{array}$

Therefore, the magnetic field is pointing towards the north and the current point to the east. The direction of the force is upwards.

If the current in the wire is running upward

Since the current in the wire is running the upward direction, therefore the magnetic field is given by:

$\begin{array}{r}\mathrm{F}=\mathrm{ILBsin}\mathrm{\varphi F}\\ =\mathrm{IIBsin}\left(90\right)\\ =\mathrm{IIB}(1.5)(2.5)\left(0.55\times {10}^{-4}\mathrm{T}\right)\\ =2.1\times {10}^{-4}\mathrm{N}\end{array}$

Therefore, the field points to the north and the current points out of the page. The direction of the force is from east to west.

Again, since the current in the wire is running from north to south in which case the angle is$180°$ , since$sin180°=0$,

Therefore, F = 0.

The magnetic force is very small so not enough to cause significant effects.