Question

A rail gun uses electromagnetic forces to accelerate a projectileto very high velocities. The basic mechanism of acceleration isrelatively simple and can be illustrated in the following example.A metal rod of mass 50.0 and electricalresistance 0.400 $\Omega$ rests on parallel horizontalrails that have negligible electric resistance. The rails are adistance = 6.00 apart. The rails are alsoconnected to a voltage source providing a voltage of = 5.00 .
The rod is placed in a vertical magnetic field. The rod begins toslide when the field reaches the value = 0.327 . Assume that the rod has a slightly flattenedbottom so that it slides instead of rolling.Use 9.80 $m/s^2$ for the magnitude of theacceleration due to gravity. Find $mu_s$ , the coefficient of staticfriction between the rod and the rails.   

I used theformula μs= VBL/Rmg and got 5.005*10^-4 nut that is not theanswer.

Answer 1

Concepts and reason

The concepts used to solve this problem are magnetic force and maximum static friction.

First, calculate magnetic force acting on the rod.

Then, calculate the maximum static friction force acting on the rod.

Finally, equate the both expressions of force and rearrange it for coefficient of static friction.

Fundamentals

The expression for magnetic force acting on a the current carrying wire is given by,

$F = BIL\sin \theta$

Here $B$ is magnetic field strength, $L$ is length of wire, $I$ is current in the wire and $\theta$ is angle between $B$ and $L$.

The expression of maximum static friction force acting on an object is given by,

${f_s} = {\mu _s}mg$

Here ${\mu _s}$ is coefficient of static friction, $m$ is mass and $g$ is acceleration due to gravity.

The expression for magnetic force acting on a the current carrying rod is given by,

$F = BIL\sin \theta$

Here $B$ is magnetic field strength, $L$ is length of rod, $I$ is current in the rod and $\theta$ is angle between $B$ and $L$.

The current in the rod is calculated by,

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

Here $V$ is potential difference and $R$ is resistance.

Substitute $5.00{\rm{ V}}$ for $V$ and $0.400{\rm{ }}\Omega$ for $R$,

$\begin{array}{c}\\I = \frac{{5.00{\rm{ V}}}}{{0.400{\rm{ }}\Omega }}\\\\ = 12.5{\rm{ A}}\\\end{array}$

Substitute $0.327{\rm{ T}}$ for $B$, $6.00{\rm{ cm}}$ for $L$, $12.5{\rm{ A}}$ for $I$ and $90^\circ$ for $\theta$ in $F = BIL\sin \theta$,

$\begin{array}{c}\\F = \left( {0.327{\rm{ T}}} \right)\left( {12.5{\rm{ A}}} \right)\left( {6.00{\rm{ cm}}} \right)\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)\sin 90^\circ \\\\ = 0.245{\rm{ N}}\\\end{array}$

The expression of maximum static friction force acting on rod is given by,

${f_s} = {\mu _s}mg$

Here ${\mu _s}$ is coefficient of static friction, $m$ is mass and $g$ is acceleration due to gravity.

Equate this force with the magnetic force.

$\begin{array}{c}\\{f_s} = F\\\\{\mu _s}mg = 0.245{\rm{ N}}\\\end{array}$

Rearrange ${\mu _s}mg = 0.245{\rm{ N}}$ to solve for ${\mu _s}$.

${\mu _s} = \frac{{0.245{\rm{ N}}}}{{mg}}$

Substitute $50.0{\rm{ g}}$ for $m$ and $9.80{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ in the above equation.

$\begin{array}{c}\\{\mu _s} = \frac{{0.245{\rm{ N}}}}{{\left( {50.0{\rm{ g}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)\left( {9.80{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)}}\\\\ = 0.500\\\end{array}$

Ans:

The coefficient of static friction between rod and the rails is $0.500$.