91Ó°ÊÓ

Magnetic flux is given by

$\mathit{ϕ}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{A}}$

Where$\stackrel{\mathbf{→}}{\mathbf{B}}$ is the magnetic field andis the area vector

Magnetic Torque is given by

$\stackrel{\mathbf{→}}{\mathbf{τ}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{μ}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\mathit{I}\mathit{A}\mathit{B}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{ϕ}$

Faraday’s law states that The induced emf in a coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil. It involves the interaction of charge with magnetic field.

$\mathit{e}\mathit{m}\mathit{f}\mathbf{=}\mathbf{-}\mathit{N}\frac{\mathbf{∆}\mathbf{ϕ}}{\mathbf{∆}\mathbf{t}}$

Where$\mathit{Ï•}$ is the angle between the area vector and Magnetic Field.

Moment of inertia of a rod about an axis through its end is given by

$\mathit{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathit{M}{\mathit{L}}^{\mathbf{2}}$

When the loop is left to rotate about the x-axis, the weight of the other sides exert torque on it

$\begin{array}{c}{\mathrm{τ}}_g=\frac{m}{4}g\frac{L}{2}\sin (90−\mathrm{ϕ})+\frac{m}{4}gL\sin (90−\mathrm{ϕ})+\frac{m}{4}g\frac{L}{2}\sin (90−\mathrm{ϕ})\\ {\mathrm{τ}}_g=\frac{mgL}{2}\cos \mathrm{ϕ}\end{array}$

The direction of the rotation is clockwise

${\stackrel{→}{\mathrm{τ}}}_g=\frac{mgL}{2}\cos \mathrm{ϕ}$

Flux change is given by

$\frac{d\mathrm{φ}}{dt}=-BA\sin \mathrm{Ï•}\frac{d\mathrm{Ï•}}{dt}=-BA\mathrm{Ó\neg }\sin \mathrm{Ï•}$

According to faraday’s laws

$\mathrm{Î\mu }=-\frac{d\mathrm{φ}}{m}=BA\mathrm{Ó\neg }\sin \mathrm{Ï•}$

Since, the emf is positive, using the right hand rule, direction of induced current is counter clockwise

Since the loop carries current and moves, in the magnetic field , so thereis a magnetic force on it exerting a torque.

${\mathrm{Ï„}}_B=IBAsin\mathrm{Ï•}$

Using right-hand rule for a cross product, the direction of this torque iis counter clockwise.

From Ohm’s law

$\begin{array}{r}I=\frac{\mathrm{Î\mu }}{R}=\frac{BA\mathrm{Ó\neg }\sin \mathrm{Ï•}}{R}\\ {\mathrm{Ï„}}_B=\frac{BA\mathrm{Ó\neg }\sin \mathrm{Ï•}}{R}BA\sin \mathrm{Ï•}\\ {\mathrm{Ï„}}_B=\frac{B^2{A}^2\mathrm{Ó\neg }}{R}\sin \mathrm{Ï•}\end{array}$

Therefore the net torque is given by

$\begin{array}{c}{\stackrel{→}{\mathrm{Ï„}}}_n={\stackrel{→}{\mathrm{Ï„}}}_g+{\stackrel{→}{\mathrm{Ï„}}}_B\\ {\stackrel{→}{\mathrm{Ï„}}}_n=−\frac{mgL}{2}\cos \mathrm{Ï•}+\frac{B^2{L}^4\mathrm{Ó\neg }}{R}\sin \mathrm{Ï•}\end{array}$

We know that

$T=l\mathrm{Î\pm }$

Moment of inertia of the loop is

$$\begin{gathered} I=\frac{1}{3}\frac{m}{4}{L}^2+\frac{m}{4}{L}^2+\frac{1}{3}\frac{m}{4}{L}^2 \\ l=\frac{2}{5}m{L}^2 \end{gathered}$$

So the angular acceleration is given by

$\begin{array}{c}\mathrm{Î\pm }=\frac{\mathrm{Ï„}}{1}\\ \mathrm{Î\pm }=\frac{12B^2{L}^2\mathrm{Ó\neg }}{5mR}{\sin }^2\mathrm{Ï•}−\frac{6g}{5L}\cos \mathrm{Ï•}\end{array}$

The magnetic torque slows down the dall since it opposes the gravitational torque, So it takes the loop a longer time to rotate through 90 degrees

No, the energy is not conserved because some of the energy is dissipated in the resistance of the loop.