91Ó°ÊÓ

Length of the wire, L = 2.0m

Angle made by the wire with x-axis,$\mathrm{θ}={60}^o$

Current carried by the wire, i = 2.0 A

The problem deals with the simple calculation of magnetic force for different magnetic fields. A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials.

Formula:

Magnetic force$F_B=iLB\sin \mathrm{θ}$

The magnetic force is given by

$F_B=iLB\sin \mathrm{θ}$

Where

i = Current, L = Length of the conductor, B = Magnetic field,

$\mathrm{θ}$= Angle between current and field

The magnetic field is $\left(4.0\hat{k}\right)T$.

Since it is perpendicular to the wire, i.e., $\mathrm{θ}={90}^o$, the magnetic force is given by

F_B = (2.0A) (2.0m) (4.0T) (sin90^o)

If we resolve the section of wire along x and y axes, by symmetry, we can say that the component along x axis gets vanished.

The net resultant is only because of y components.

$\begin{array}{rcl}{\stackrel{â‡\P Ä}{F}}_{net}& =& 2F\sin \left({90}^o-{60}^o\right)\\ & =& 2×16N×\sin {30}^o\\ {\stackrel{â‡\P Ä}{F}}_{net}& =& \left(-16.0N\right)\hat{j}\end{array}$

Now magnetic field is$\left(4.0\hat{i}\right)T$, that is, along x axis.

By the right-hand rule, the force exerted on the left half of the bent wire points in the role="math" localid="1662895688333" $-\hat{k}$direction, and the force exerted on the right half of the wire points in the$+\hat{k}$direction. It is clear that the magnitude of each force is equal so that the force evaluated over the entire wire must necessarily vanish.

Thus, the magnitude of magnetic force is zero.