91Ó°ÊÓ

The mass of the metal bar is m .

The distance between the two rails is I.

The resistance across the rails is R.

The equation to calculate the (According to Faraday’s law) magnetic flux is given as follows.

$\mathbf{ϕ}\mathbf{=}\mathbf{∫}\mathbf{B}\mathbf{-}\mathbf{dA}$ …… (1)

Here, B is the magnetic field and A is the area of the surface.

The equation to calculate the magnitude of induced emf (According to Faraday’s law of induction) is given as follows.

$\mathbf{Î\mu }\mathbf{=}\mathbf{-}\frac{\mathbf{»åÏ•}}{\mathbf{dt}}$ …… (2)

The equation to calculate the current in the resistor is given as follows.

$\mathbf{i}\mathbf{=}\frac{\mathbf{Î\mu }}{\mathbf{R}}$ …… (3)

The equation to calculate the magnetic force on the bar is given as follows.

$\mathbf{F}\mathbf{=}\mathbf{iIB}$ …… (4)

Here, is the current.

The equation to calculate the energy delivered to the resistor is given as follows.

$\mathbf{ϕ}\mathbf{=}\mathbf{∫}\mathbf{B}\mathbf{-}\mathbf{dA}\mathbf{}\mathbf{}\mathbf{W}\mathbf{=}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{∞}}{\mathbf{i}}^{\mathbf{2}}\mathbf{Rdt}$ …… (5)

(a)

Consider the diagram that shows the slide of the bar on the frictionless parallel rails.

Let assume the bar moves along the x-axis and field along the negative y-axis.

The magnetic field is given by,

$B=-B\hat{y}$

The area vector with area of the loop as ,

$da=day$

Calculate the magnetic flux

Substitute$-B\hat{y}$for B and day for into equation (1).

$$\begin{gathered} \mathrm{ϕ}=∫\left(B\hat{y}\right).\left(day\right) \\ \mathrm{ϕ}=B∫da \end{gathered}$$

$\mathrm{Ï•}=BA$

Calculate the induced emf.

Substitute -BA for $\mathrm{Ï•}$into equation (2).

$$\begin{gathered} \mathrm{Î\mu }=-\frac{d}{dt}\left(-BA\right) \\ \mathrm{Î\mu }=-B\frac{dA}{dt} \end{gathered}$$

Substitute Ix for A into above equation.

$$\begin{gathered} \mathrm{Î\mu }=-B\frac{d}{dt}\left(Ix\right) \\ \mathrm{Î\mu }=-BI\frac{dx}{dt} \end{gathered}$$

Substitute v for$\frac{dx}{dt}$ into above equation.

$\mathrm{Î\mu }=-BIv$

Calculate the current in the resistor.

Substitute B/v for $\mathrm{Î\mu }$into equation (3).

$i=\frac{BIv}{R}$

Hence the current in the resistor is $\frac{BIv}{R}$and direction of the current is counter clockwise.

(b)

Calculate the magnetic force on the bar.

Substitute $\frac{BIv}{R}$for i into equation (4).

$$\begin{gathered} F=\frac{BIv}{R}IB \\ F=\frac{B^2{I}^{2}v}{R} \end{gathered}$$

Hence the magnitude of the magnetic force is$\frac{B^2{I}^{2}v}{R}$and its direction is negative x-axis

(c)

Now the slide of the bar is lift side, therefore the magnetic force after time ,

$F=-\frac{B^2{I}^{2}v}{R}$ …… (6)

According to the second law of newtons, the force on the bar,

$\mathrm{F}=\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}$ …… (7)

Now equate the equation (6) and (7),

$$\begin{gathered} m\frac{dv}{dt}=-\frac{B^2{I}^{2}v}{R} \\ \frac{dv}{v}=\frac{B^2{I}^2}{Rm}dt \end{gathered}$$

Integrate the above equation.

${∫}_{v_0}^v\frac{dv}{v}=-{∫}_0^t\frac{B^2{I}^2}{Rm}dt$

$$\begin{gathered} \mathrm{In}\left(\mathrm{v}\right)-\mathrm{In}\left({\mathrm{v}}_0\right)=-\frac{{\mathrm{B}}^2{\mathrm{I}}^2}{\mathrm{Rm}}\left(\mathrm{t}\right) \\ \mathrm{In}\left(\frac{\mathrm{v}}{{\mathrm{v}}_0}\right)=-\frac{{\mathrm{B}}^2{\mathrm{I}}^2}{\mathrm{Rm}}\mathrm{t} \\ \frac{\mathrm{v}}{{\mathrm{v}}_0}={\mathrm{e}}^{\frac{{\mathrm{B}}^2{\mathrm{I}}^2}{\mathrm{Rm}}\mathrm{t}} \\ \mathrm{Solve}\mathrm{further}\mathrm{as}, \\ \mathrm{v}={\mathrm{v}}_0{\mathrm{e}}^{\frac{{\mathrm{B}}^2{\mathrm{I}}^2}{\mathrm{Rm}}\mathrm{t}} \\ \mathrm{Hence}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{bar}\mathrm{after}\mathrm{time}\mathrm{t}\mathrm{is}{\mathrm{v}}_0{\mathrm{e}}^{\frac{{\mathrm{B}}^2{\mathrm{I}}^2}{\mathrm{Rm}}\mathrm{t}}. \end{gathered}$$

(d)

Calculate the energy delivered to resistor.

Substitute $\frac{BIv}{R}$for i into equation (5).

$W={∫}_0^{∞}{\left(\frac{BIv}{R}\right)}^{2}Rdt$

Substitute $v_0{e}^{-\frac{B^2{I}^{2}v}{Rm}t}$for v into above equation.

$$\begin{gathered} W={∫}_0^{∞}{\left({}^{-\frac{BI}{R}{v}_0{e}^{{}^{-\frac{B^2{I}^{2}v}{Rm}t}}}\right)}^{2}Rdt \\ W={\left(\frac{BI{v}_0}{R}\right)}^2{R}_0{∫}_0^{∞}\left(e^{{}^{{}^{-\frac{2B^2{I}^{2}v}{Rm}t}}}\right)dt \\ W=\frac{{\left(BI{v}_0\right)}^2}{R}{\left(-\frac{Rm}{2B^2{i}^2}{e}^{{}^{{}^{-\frac{2B^2{I}^{2}v}{Rm}t}}}\right)}_0^{∞} \\ W=-\frac{{\left(BI{v}_0\right)}^2}{R}×\frac{Rm}{2B^2{I}^2}{\left(e^{{}^{-\frac{2B^2{I}^{2}v}{Rm}t}}\right)}_0^{∞} \end{gathered}$$

$$\begin{gathered} \mathrm{Apply}\mathrm{the}\mathrm{limits}, \\ W=-\frac{{\left(BI{v}_0\right)}^2}{R}×\frac{Rm}{2B^2{I}^2}\left(e^{-\frac{2B^2{I}^2}{Rm}\left(∞\right)}{e}^{-\frac{2B^2{I}^2}{Rm}\left(0\right)}\right) \\ W=-\frac{{\left(BI{v}_0\right)}^2}{R}×\frac{Rm}{2B^2{I}^2}\left(0-1\right) \\ W=V_0^2×\frac{m}{2} \\ W=\frac{1}{2}m \\ \mathrm{Hence}\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{that}\mathrm{the}\mathrm{energy}\mathrm{delivered}\mathrm{to}\mathrm{the}\mathrm{resistor}\mathrm{is}\frac{1}{2}{\mathrm{mv}}_0^2. \end{gathered}$$