91Ó°ÊÓ

Magnetic flux is given by

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

Where$\stackrel{â‡\P Ä}{B}$is the magnetic field and$\stackrel{→}{A}$is the area vector

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.

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

Force on conductor with vector$\stackrel{â‡\P Ä}{I}$ along the wire and placed in magnetic field$\stackrel{→}{B}$ is

${\stackrel{→}{F}}_b=I\stackrel{→}{l}×\stackrel{→}{B}$

Power delivered to a resistor of resistance R is given by

$\mathit{P}\mathbf{=}{\mathit{I}}^{\mathbf{2}}\mathit{R}$

Apply the formula for flux change

$\mathrm{ϕ}=\stackrel{→}{B}·\stackrel{→}{A}=BA\cos \left(\mathrm{θ}\right)$

As the bar slides down, the area of the loop increases

$$\begin{gathered} \frac{d\mathrm{ϕ}}{dt}=B\cos \left(\mathrm{θ}\right)\frac{dA}{dt}=BL\cos \left(\mathrm{θ}\right)\frac{ds}{dt} \\ \frac{d\mathrm{ϕ}}{dt}=BLv\cos \left(\mathrm{θ}\right) \end{gathered}$$

Using faraday’s law we get

$\mathrm{Î\mu }=-BLv\cos \left(\mathrm{θ}\right)$

So, induced emf is negative.

Using the right-hand rule, induced emf is directed counter clockwise

Therefore the induced current is a to b.

Since the bar is a current carrying conductor, moving in a magnetic field, so there is magnetic force on the bar is

$F_B=ILB$

From Ohm’s law

$F_B=\frac{LB\mathrm{Î\mu }}{R}$

Substitution for the emf here we get

$F_B=\frac{L^2{B}^{2}v}{R}\cos \left(\mathrm{Ï•}\right)$

Since the bar slides down due to its own weight we get

$F_g=mg\sin \left(\mathrm{Ï•}\right)$

When terminal velocity is reached we get equilibrium

$$\begin{gathered} F_g=F_B \\ mg\sin \left(\mathrm{Ï•}\right)=\frac{L^2{B}^2{v}_t}{R}\cos \left(\mathrm{Ï•}\right) \end{gathered}$$

Rearranging we get terminal velocity is

$v_t=\frac{mgR\sin \mathrm{Ï•}}{L^2{B}^2{\cos }^2\mathrm{Ï•}}$

We know at terminal velocity emf is

$\mathrm{Î\mu }=BL{v}_t\cos \mathrm{Ï•}$

Again from Ohm’s Law

$I=\frac{\mathrm{Î\mu }}{R}=\frac{BL{v}_t\cos \mathrm{Ï•}}{R}$

Substitute the expression for the terminal velocity

$I=\frac{mg\tan \mathrm{Ï•}}{LB}$

Power delivered is given by

$P_e=I^{2}R$

Substitute the expression for current at terminal velocity

$P_e=\frac{m^2{g}^{2}R{\tan }^2\mathrm{Ï•}}{L^2{B}^2}$

Work done is given by

$W=F.s$

Rate of work at which work is done

$$\begin{gathered} \frac{d{W}_g}{dt}=mg\sin \mathrm{Ï•}\frac{ds}{dt} \\ \frac{d{W}_g}{dt}=mg{v}_t\sin \mathrm{Ï•} \end{gathered}$$

Substitute the expression for terminal velocity

$\frac{d{W}_g}{dt}=\frac{m^2{g}^{2}R{\tan }^2\mathrm{Ï•}}{L^2{B}^2}$

This is the same as power in part (e) as in by Conversation of energy.