Chapter 31, Problem 31.83CP

Review. The bar of mass m in Figure P30.51 is pulled horizontally across parallel, frictionless rails by a massless string that passes over a light, frictionless pulley and is attached to a suspended object of mass M. The uniform upward magnetic field has a magnitude B, and the distance between the rails is ℓ. The only significant electrical resistance is the load resistor R shown connecting the rails at one end. Assuming the suspended object is released with the bar at rest at t = 0, derive an expression that gives the bar’s horizontal speed as a function of time.

Figure P30.51

To determine

The expression for horizontal speed of the bar as a function of time.

Answer to Problem 31.83CP

The expression for horizontal speed of the bar as a function of time is $\frac{MgR}{B^2{l}^2}\left(1-e^{-\frac{B^2{l}^{2}t}{\left(m+M\right)R}}\right)$.

Explanation of Solution

Given info: Mass of bar is $m$, mass of object is $M$, upward magnetic field is $B$, distance between the rails is $l$ and resistance of load resistor is $R$.

The emf induced in the bar can be given as,

$\epsilon =Blv$

Here,

$\epsilon$ is the emf induced in the bar.

$B$ is the uniform magnetic field

$l$ is the distance between the rails.

$v$ is the horizontal speed of bar.

The current induced in the bar can be given as,

$I=\frac{\epsilon }{R}$

Here,

$I$ is the current induced in the bar.

$R$ is the resistance of the load resistor.

Substitute $Blv$ for $I$ in the above equation,

$I=\frac{Blv}{R}$

The force induced in the bar due to magnetic field can be given as,

$F=IBl$

Here,

$F$ is the force induced in the bar.

Substitute $\frac{Blv}{R}$ for $I$ in the above equation,

$\begin{array}{c}F=\left(\frac{Blv}{R}\right)Bl\\ =\frac{B^2{l}^{2}v}{R}\end{array}$ (1)

The force due to weight can be given as,

$W=Mg$ (2)

Here,

$M$ is the mass of object.

$g$ is the acceleration due to gravity.

$W$ is the force due to weight.

As, force due to magnetic field and force due to weight will act in opposite direction, the net force acting on the bar can be given by subtracting equation (2) from equation (1),

$F_{\text{net}}=Mg-\frac{B^2{l}^{2}v}{R}$ (3)

Here,

$F_{\text{net}}$ is the net force acting on the bar.

The net force can also be given as,

$F_{\text{net}}=\left(m+M\right)\frac{dv}{dt}$

Substitute $\left(m+M\right)\frac{dv}{dt}$ for $F_{\text{net}}$ in the equation (3),

$\begin{array}{c}\left(m+M\right)\frac{dv}{dt}=Mg-\frac{B^2{l}^{2}v}{R}\\ \frac{dv}{dt}+\frac{B^2{l}^{2}v}{\left(m+M\right)R}=\frac{M}{\left(m+M\right)}g\end{array}$ (4)

The equation (4) is a linear differential equation of the form,

$\frac{dv}{dt}+Pv=Q$

Here,

$P$ is $\frac{B^2{l}^2}{\left(m+M\right)R}$ and a function of $t$.

$Q$ is $\frac{M}{\left(m+M\right)}g$ and a function of $t$.

The integrating factor for the equation (4) can be given as,

$\begin{array}{c}I.F.=e^{{\displaystyle \int Pdt}}\\ =e^{{\displaystyle \int \frac{B^2{l}^2}{\left(m+M\right)R}dt}}\\ =e^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}\end{array}$

Here,

$I.F.$ is the integrating factor.

The solution for the differential equation is,

$v(I.F.)={\displaystyle \int Q(I.F.)dt}+C$

Here,

$C$ is the constant.

Substitute $e^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}$ for $I.F.$ in the above equation,

$\begin{array}{c}v{e}^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}={\displaystyle \int \frac{M}{\left(m+M\right)}g{e}^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}dt}+C\\ v{e}^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}=\left(\frac{M}{\left(m+M\right)}\right)\left(\frac{\left(m+M\right)R}{B^2{l}^2}\right)g{e}^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}+C\\ v{e}^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}=\frac{MgR}{B^2{l}^2}{e}^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}+C\end{array}$ (5)

Apply boundary condition, $v=0$ at $t=0$ in the above equation,

$\begin{array}{c}0=\frac{MgR}{B^2{l}^2}+C\\ C=-\frac{MgR}{B^2{l}^2}\end{array}$

Substitute $\left(-\frac{MgR}{B^2{l}^2}\right)$ for $C$ in the equation (5),

$\begin{array}{l}v{e}^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}=\frac{MgR}{B^2{l}^2}{e}^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}-\frac{MgR}{B^2{l}^2}\\ v{e}^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}=\frac{MgR}{B^2{l}^2}\left(e^{\frac{B^2{l}^{2}t}{\left(m+M\right)R}}-1\right)\\ v=\frac{MgR}{B^2{l}^2}\left(1-e^{-\frac{B^2{l}^{2}t}{\left(m+M\right)R}}\right)\end{array}$

Thus, the expression for speed of the bar is $\frac{MgR}{B^2{l}^2}\left(1-e^{-\frac{B^2{l}^{2}t}{\left(m+M\right)R}}\right)$.

Conclusion:

Therefore, the expression for horizontal speed of the bar as a function of time will be $\frac{MgR}{B^2{l}^2}\left(1-e^{-\frac{B^2{l}^{2}t}{\left(m+M\right)R}}\right)$.