A single-turn current loop, carrying a current of4.00 A ,is in the shape of a right triangle with sides 50.0, 120, and 130cm. The loop is in a uniform magnetic field of magnitude 75.0mT whose direction is parallel to the current in the 130cm side of the loop. (a) What is the magnitude of the magnetic force on the 130cm side? (b) What is the magnitude of the magnetic force on the 50.0cm side? (c) What is the magnitude of the magnetic force on the 120cmside? (d)What is the magnitude of the net force on the loop?

1. Current carried by the single turn loop, i = 4.00 A
2. The lengths of the sides of a right angled triangle,
   $l_x=120cm,l_y=50cm,l_z=130cm$
   
3. The uniform, magnetic field of the loop, B = 75.0mT
4. The direction of magnetic field is parallel to the current in the 130 cm side of the loop.

If a particle is moving with a uniform velocity within a uniform magnetic field, then it experiences a magnetic force due to its charge value that induces a current within the loop. The force acting on the particle is due to the current along the length of the conductive wire. The direction of this magnetic force is given by Fleming's right-hand rule as the force is perpendicular to both the speed and magnetic field acting on it.

The magnetic force along a loop wire is given as:

$F_B=ilB\sin \theta$ …… (i)

Here, i is the current in the loop, B is the magnetic field that it experiences, l is the length of the conducting wire, $\theta$is the angle made by the length vector with the magnetic field.

The angle made by a vector along an axis is as follows:

$\theta ={\tan }^{-1}\left(\frac{Perpendicular}{Height}\right)$ …… (ii)

The magnitude of a vector is as follows:

$\left|A\right|=\sqrt{A_x^2+A_y^2}$ ….. (iii)

Here, A_x is the horizontal component of the vector, A_y is the vertical component of the vector.

Consider the direction of magnetic field is parallel to the current in the 130 cm side of the loop. That implies, the angle between them will be $\theta =0^o$.

Thus, the magnitude of the magnetic field on the side 130 cm is given using equation (i) as follows:

$\begin{array}{rcl}{F}_B& =& ilB\sin \left(0^o\right)\\ & =& 0N\end{array}$

Hence, the value of the force is 0 N.

The angle made by hypotenuse with -x axis is given using equation (ii) as follows:

$\begin{array}{rcl}\varphi & =& {\tan }^{-1}\left(\frac{50cm}{120cm}\right)\\ & =& 22.6^o\end{array}$

Thus, the angle made by hypotenuse with +x -axis is given by:

$\begin{array}{rcl}\theta & =& 180-22.6^o\\ & & ~{157}^o\end{array}$

Now, the horizontal and vertical components of the magnetic field can be given as follows:

$\begin{array}{rcl}{B}_x& =& -B\cos \theta \\ & =& -\left(0.075T\right)\cos \left(22.6^o\right)\\ & =& -0.0692T\end{array}$

Consider the y component as:

$\begin{array}{rcl}{B}_y& =& B\sin \theta \\ & =& \left(0.075T\right)\sin \left(22.6^0\right)\\ & =& 0.0288T\end{array}$

Thus, the magnitude of the magnetic field can be given using equation (iii) as follows:

$\begin{array}{rcl}\left|B\right|& =& \sqrt{{\left(-0.0692T\right)}^2+{\left(0.0288T\right)}^2}\\ & =& 0.0692T\end{array}$

The magnitude of magnetic force on the 50.0 cm side is given by equation (i) as follows:

$\begin{array}{rcl}{F}_B& =& 4.00A\times 0.5m\times 0.0692T\\ & =& 0.138N\end{array}$

Hence, the value of the magnetic force is 0.138N.

Here, the magnitude of magnetic field acting on the side is given by its perpendicular vector that is calculated in part (b) as: B_y= 0.0288 T

Similarly, the magnitude of magnetic force on the 1230 cm side is given using the data in equation (i) as follows:

$\begin{array}{rcl}{F}_B& =& 4.00A\times 1.20m\times 0.0288T\\ & =& 0.138N\end{array}$

Hence, the value of the magnetic force is 0.138 N.

The net force on the loop is zero since the directions of force on the sides of the loop are opposite.

So, the net force acting in the loop for the given current flow is given by using equation (i) as follows:

$\begin{array}{rcl}\stackrel{\rightharpoonup }{F_{net}}& =& i{l}_y{B}_x\hat{k}+i{l}_x{B}_y\hat{k}\\ & =& 0N\end{array}$

Hence, the net force value is 0 N.