91影视

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$\mathrm{桅}=90掳$

Hence the magnitude of the force is given by:

$\begin{array}{r}\mathrm{F}=\mathrm{ILBsin}鈦�\mathrm{胃}\\ =\mathrm{IIBsin}鈦�\left({90}^{鈭�}\right)\\ =\mathrm{IIB}\\ =(1.5\mathrm{A})(2.5\mathrm{m})\left(0.55脳{10}^{鈭�4}\right)\\ =2.1脳{10}^{鈭�4}\mathrm{N}\end{array}$

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

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

$\begin{array}{r}\mathrm{F}=\mathrm{ILBsin}鈦�\mathrm{蠒贵}\\ =\mathrm{IIBsin}鈦�\left(90\right)\\ =\mathrm{IIB}(1.5)(2.5)\left(0.55脳{10}^{鈭�4}\mathrm{T}\right)\\ =2.1脳{10}^{鈭�4}\mathrm{N}\end{array}$

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$180掳$ , since$sin180掳=0$,

Therefore, F = 0.

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