91Ó°ÊÓ

The term magnetic field may be defined as the area around the magnet behave like a magnet.

The magnitude of the magnetic field in that case

$\begin{array}{r}\mathrm{B}=\sqrt{{\left(\frac{{\mathrm{B}}_0\mathrm{z}}{\mathrm{L}}\right)}^2+{\left(\frac{{\mathrm{B}}_0\mathrm{y}}{\mathrm{L}}\right)}^2}\\ \mathrm{B}=\left(\frac{{\mathrm{B}}_0}{\mathrm{L}}\right)\sqrt{{\mathrm{z}}^2+{\mathrm{y}}^2}\\ \frac{\mathrm{BL}}{{\mathrm{B}}_0}=\sqrt{{\mathrm{z}}^2+{\mathrm{y}}^2}\end{array}$

Using this equation first fixed one variable and rotate other and again fixed other and rotate first and use the python code for sketch magnetic lines diagram

The diagram of magnetic field lines is

Now

For side (0,0) - (0,L) the force is

$\begin{array}{r}\stackrel{→}{\mathrm{F}}={∫}_0^{\mathrm{L}} \mathrm{Id}\stackrel{→}{\mathrm{I}}×\stackrel{→}{\mathrm{B}}\\ \stackrel{→}{\mathrm{F}}=âˆ\pounds {∫}_0^{\mathrm{L}} \frac{{\mathrm{B}}_0\mathrm{ydy}}{\mathrm{L}}\widehat{\mathrm{i}}\\ \stackrel{→}{\mathrm{F}}=\frac{1}{2}{\mathrm{B}}_0\mathrm{Ll}\widehat{\mathrm{i}}\end{array}$

For side (0,L) (L,L) the force is

$\begin{array}{r}\stackrel{→}{\mathrm{F}}={∫}_{0,\mathrm{y}=\mathrm{L}}^{\mathrm{L}} âˆ\pounds \mathrm{d}\stackrel{→}{\mathrm{I}}×\stackrel{→}{\mathrm{B}}\\ \stackrel{→}{\mathrm{F}}=âˆ\pounds {∫}_0^{\mathrm{L}} \frac{{\mathrm{B}}_0\mathrm{ydx}}{\mathrm{L}}\widehat{\mathrm{j}}\\ \stackrel{→}{\mathrm{F}}=−{\mathrm{B}}_0\mathrm{L}âˆ\pounds \widehat{\mathrm{j}}\end{array}$

For (L,L) - (L,0) the force is

$\begin{array}{r}\stackrel{→}{\mathrm{F}}={∫}_{0,\mathrm{x}=\mathrm{L}}^{\mathrm{L}} âˆ\pounds \mathrm{d}\stackrel{→}{\mathrm{I}}×\stackrel{→}{\mathrm{B}}\\ \stackrel{→}{\mathrm{F}}=\mathrm{I}{∫}_{0,\mathrm{x}=\mathrm{L}}^{\mathrm{L}} \frac{{\mathrm{B}}_0\mathrm{ydy}}{\mathrm{L}}(−\widehat{\mathrm{i}})\\ \stackrel{→}{\mathrm{F}}=−\frac{1}{2}{\mathrm{B}}_0\mathrm{L}âˆ\pounds \widehat{\mathrm{i}}\end{array}$

For (L,0) - (0,0) the force is

$\begin{array}{r}\stackrel{→}{\mathrm{F}}={∫}_{\mathrm{L},\mathrm{y}=0}^0 \mathrm{Id}\stackrel{→}{\mathrm{I}}×\stackrel{→}{\mathrm{B}}\\ \stackrel{→}{\mathrm{F}}=\mathrm{I}{∫}_{\mathrm{L},\mathrm{y}=0}^0 \frac{{\mathrm{B}}_0\mathrm{ydx}}{\mathrm{L}}\widehat{\mathrm{j}}\\ \stackrel{→}{\mathrm{F}}=0\end{array}$

Hence, the magnitude and direction of the magnetic force exerted on each of the sides of the loop by integrating is$\mathrm{d}\stackrel{→}{\mathrm{F}}=\mathrm{Id}âˆ\pounds \stackrel{→}{\mathrm{I}}×\stackrel{→}{\mathrm{B}}\text{is}\stackrel{→}{\mathrm{F}}=\frac{1}{2}{\mathrm{B}}_0\mathrm{L}|\widehat{\mathrm{i}},\stackrel{→}{\mathrm{F}}=−|{\mathrm{B}}_0\mathrm{L}\widehat{\mathrm{j}},\stackrel{→}{\mathrm{F}}=−\frac{1}{2}{\mathrm{IB}}_0\mathrm{L}\widehat{\mathrm{L}},\stackrel{→}{\mathrm{F}}=0$.

Now

The magnitude and direction of the net magnetic force on the loop is the sum of all forces${\stackrel{→}{\mathrm{F}}}_{\text{total}}=−âˆ\pounds {\mathrm{B}}_0\mathrm{L}\widehat{\mathrm{j}}$