Question

The bent wire shown in Figure lies in a uniform magnetic field. Each straight section is 2.0 m long and makes an angle of $\mathit{\theta }\mathbf{=}{\mathbf{60}}^{\mathbf{o}}$with the xaxis, and the wire carries a current of 2.0A. (a) What is the net magnetic force on the wire in unit vector notation if the magnetic field is given by $\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$ T? (b) What is the net magnetic force on the wire in unit vector notation if the magnetic field is given by $\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}$T?

Short answer

1. The net magnitic force is on the wire in unit vector notation if the magnetic field is $4\hat{k}T$is$\left(-16.0N\right)\hat{j}$
2. The net magnetic force is on the wire in unit vector notation if the magnetic field is $4\hat{i}T$is zero

Step-by-step solution

Listing the given quantities

Length of the wire, L = 2.0m

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

Current carried by the wire, i = 2.0 A

Step 2: To understand the concept

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$F_B=iLB\sin \theta$

(a) To find the net magnitic force is on the wire in unit vector notation if the magnetic field is 4k^ T

The magnetic force is given by

$F_B=iLB\sin \theta$

Where

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

$\theta$= 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., $\theta ={90}^o$, 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.

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

(b) To find the net magnitic force is on the wire in unit vector notation if the magnetic field is 4i^ T 

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.