The force on the rectangular loop of wire in the magnetic field in Figure 22.56 can be used to measure field strength. The field is uniform, and the plane of the loop is perpendicular to the field. (a) What is the direction of the magnetic force on the loop? Justify the claim that the forces on the sides of the loop are equal and opposite, independent of how much of the loop is in the field and do not affect the net force on the loop. (b) If a current of 5.00 A is used, what is the force per tesla on the 20.0-cm-wide loop?

Figure 22.56 A rectangular loop of wire carrying a current is perpendicular to a magnetic field. The field is uniform in the region shown and is zero outside that region.

Magnetic force is a result of electromagnetic force and is caused by the motion of charged particles. The magnetic force on a current-carrying conductor can be determined using the right-hand rule-1 such that, if fingers point towards the magnetic field and the thumb is in the direction of current then the palm will not be in the direction of magnetic force.

Quantitatively magnetic force can be estimated using the equation,

$\mathit{\theta }\mathbf{)}\mathbf{=}\mathbf{ILBsin}$………………(1)

Look at each of the three sides separately to determine the direction of the magnetic force on the loop. Let's assume that the length of the left and right sides is ${\mathrm{L}}_1$and the length of the bottom side is $L_2$.

The amount of the force will be determined using the equation (1), which will give,

${\mathrm{F}}_{\mathrm{Left}}={\mathrm{IL}}_1\mathrm{B}$

The right-hand rule is then used to identify the force's direction. We'll align the thumb with the current, which is up, and the fingers with the magnetic field, which is out of the page, with the palm pointing to the right, which is the direction of the force on the right side of the loop.

Finally, the amount and direction of the magnetic force on the loop's bottom side are determined.

Using the equation (1), we will determine the amount of force,

${\mathrm{F}}_{\mathrm{Bottom}=}{\mathrm{IL}}_2\mathrm{B}$

The right-hand rule is then used to identify the force's direction. The palm will be pointing downwards, which is the direction of the force on the bottom side of the loop, and the thumb will be pointing in the direction of the current, which is to the right, and the fingers will be pointing in the direction of the magnetic field, which is out of the page.

The loop is now subjected to three forces:

There are two horizontal forces, ${\mathrm{F}}_{\mathrm{Left}}$to the left and ${\mathrm{F}}_{\mathrm{Right}}$to the right. They will cancel each other out as they are equal and opposite, resulting in a horizontal net force of zero.

There is just one force ${\mathrm{F}}_{\mathrm{Bottom}}$acting vertically downwards.

The loop's net force is ${\mathrm{F}}_{\mathrm{Bottom}}$ and it's heading downwards.

We'll utilize equation (1) to determine the force per tesla on the bottom side of the loop since the current and magnetic field directions are perpendicular, and solve this equation,

$$\begin{gathered} \frac{\mathrm{F}}{\mathrm{B}}=\mathrm{IL} \\ =(0.20\mathrm{m})\times (5.00\mathrm{A}) \\ =1\mathrm{N}/\mathrm{T} \end{gathered}$$

$\mathrm{Therefore},1\mathrm{N}/\mathrm{T}\mathrm{is}\mathrm{the}\mathrm{force}\mathrm{per}\mathrm{tesla}$