# A copper rod of mass m rests on two horizontal rails distance L apart and carries a current of i from one rail to the other. The coefficient of static friction between rod and rails is mu_s. What are the (a) magnitude and (b) angle(relative.....?

## A copper rod of mass $m$ rests on two horizontal rails distance $L$ apart and carries a current of $i$ from one rail to the other. The coefficient of static friction between rod and rails is ${\mu}_{s}$. What are the (a) magnitude and (b) angle(relative to the vertical) of the smallest magnetic field that puts the rod on the verge of sliding?

Jan 10, 2018

We are to find the smallest magnetic field that puts the rod of length $L$ carrying current $i$ on the verge of sliding.

The magnetic force generated by current carrying conductor can be so oriented that (1) it must just push the rod horizontally to overcome the force of friction given by ${\mu}_{s} N$ where $N$ is normal reaction, (2) in addition it may lift the rod up to reduce the normal reaction force and therefore, force of friction.

Let the copper rod be oriented along $y$-axis and current flows along $+ y$-axis and that it moves along $x$-axis.

Let the uniform magnetic field be applied so that it makes an angle $\theta$ with the vertical or $\left(\frac{\pi}{2} + \theta\right)$ with $x$-axis.
It has two components

${B}_{\downarrow} = B \cos \theta$, into the paper ......(1)
${B}_{x} = B \sin \theta$ .......(2)

Force due the magnetic field will have two components as given by Fleming's right-hand rule, ${F}_{x} \mathmr{and} {F}_{\uparrow}$ out of the paper, so that

${F}_{x} = i L {B}_{\downarrow} = i L B \cos \theta$ ......(3)
${F}_{\uparrow} = i L {B}_{x} = i L B \sin \theta$ ......(4)

We see that normal reaction becomes

$N = m g - i L B \sin \theta$ ......(5)

When the rod just starts sliding, maximum force of friction, which always opposes the motion, must be equal to component of magnetic force in the $x$-direction so that net force is zero and ythere is no acceleration in this direction.

$\therefore {\mu}_{s} \left(m g - i L B \sin \theta\right) = i L B \cos \theta$ .....(6)

Solving for magnetic field we get

$B = \frac{{\mu}_{s} m g}{i L \left({\mu}_{s} \sin \theta + \cos \theta\right)}$ ......(7)

To find out minimum magnetic field we set first differential of this expression with $\theta$ equal to $0$

$\frac{\mathrm{dB}}{d \theta} = \frac{{\mu}_{s} m g \left({\mu}_{s} \cos \theta - \sin \theta\right)}{i L {\left({\mu}_{s} \sin \theta + \cos \theta\right)}^{2}}$
$= 0$,

We get

$\theta = {\tan}^{-} 1 {\mu}_{s}$ ......(8)

To calculate ${B}_{\min}$ we insert value of $\theta$ as found from (8) above in expression (7), taking $g = 9.81 m {s}^{-} 2$.