Question

A metallic rod of length $L$ and mass $M$ is moving under the action of two unequal forces $F_1$ and $F_2$ (directed opposite to each other) acting at its ends along its length. Ignore gravity and any external magnetic field. If specific charge of electrons is $\left(e/m\right)$, then the potential difference between the ends of the rod is steady state must be:

• A. $|F_1-F_2|mL/eM$
• B. $(F_1-F_2)mL/eM$
• C. $\left[mL/eM\right]ln\left[F_1/F_2\right]$
• D. None

Answer 1

The correct option is D None

For the entire rod, assuming an acceleration $a$

$F1-F2=M\times a-----\left(1\right)$

For the same rod (therefore same acceleration)

Consider only length $x$

$F1-T=\left(\frac{M\times x}{L}\right)a.$

Plugging in $a=\left(\frac{F1-F2}{M}\right)$ from equation $1$,

$F1-T=\left(\frac{M\times x}{L}\right)\times \left(\frac{F1-F2}{M}\right)$

$T=F1-\frac{\left(F1-F2\right)\times x}{L}$

Hence, The Option $D$ is correct $None$