Question

A metal bar of mass m slides frictionlessly on two parallel conducting rails a distance l apart (Fig. 7 .17). A resistor R is connected across the rails, and a uniform magnetic field B, pointing into the page, fills the entire region.

(a) If the bar moves to the right at speed V, what is the current in the resistor? In what direction does it flow?

(b) What is the magnetic force on the bar? In what direction?

(c) If the bar starts out with speed${\mathbf{V}}_{\mathbf{0}}$at time t=0, and is left to slide, what is its speed at a later time t?

(d) The initial kinetic energy of the bar was, of course,$\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$Check that the energy delivered to the resistor is exactly $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$.

Short answer

(a) The current in the resistor is$\frac{BIv}{R}$and direction of the current is counter clockwise.

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

(c) The speed of the bar after time t is $v_0{e}^{\frac{B^2{I}^{2}v}{R}t}$.

(d) It is proved that the energy delivered to the resistor is $\frac{1}{2}m{v}_0^2$.

Step-by-step solution

Write the given data from the question.

The mass of the metal bar is m .

The distance between the two rails is I.

The resistance across the rails is R.

Determine the equation to calculate the current through the resistor and its direction, the magnetic force and its direction, speed of the bar after time t and kinetic energy of the bar.

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

$\mathbf{\varphi }\mathbf{=}\mathbf{\int }\mathbf{B}\mathbf{-}\mathbf{dA}$ …… (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.

$\mathbf{\epsilon }\mathbf{=}\mathbf{-}\frac{\mathbf{d\varphi }}{\mathbf{dt}}$ …… (2)

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

$\mathbf{i}\mathbf{=}\frac{\mathbf{\epsilon }}{\mathbf{R}}$ …… (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.

$\mathbf{\varphi }\mathbf{=}\mathbf{\int }\mathbf{B}\mathbf{-}\mathbf{dA}\mathbf{}\mathbf{}\mathbf{W}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{i}}^{\mathbf{2}}\mathbf{Rdt}$ …… (5)

Calculate the current in the resistor and its direction. 

(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).

$$\begin{gathered} \varphi =\int \left(B\hat{y}\right).\left(day\right) \\ \varphi =B\int da \end{gathered}$$

$\varphi =BA$

Calculate the induced emf.

Substitute -BA for $\varphi$into equation (2).

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

Substitute Ix for A into above equation.

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

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

$\epsilon =-BIv$

Calculate the current in the resistor.

Substitute B/v for $\epsilon$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.

Calculate the magnetic force on the bar and its direction.

(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

Calculate the speed of the bar after time t.  

(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.

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

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

Calculate the initial kinetic energy of the bar.

(d)

Calculate the energy delivered to resistor.

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

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

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

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

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