A thin, 50.0-cm-long metal bar with mass 750 g rests on, but is not attached to, two metallic supports in a uniform 0.450-T magnetic field, as shown in Fig. E27.37. A battery and a 25.0-Ω resistor in series are connected to the supports. (a) What is the highest voltage the battery can have without breaking the circuit at the supports? (b) The battery voltage has the maximum value calculated in part (a). If the resistor suddenly gets partially short-circuited, decreasing its resistance to 2.0 Ω, find the initial acceleration of the bar.

Given,

Length of the bar is l = 0.500m , mass m = 0.750kg , the magnetic field is B=0.450T and the resistance is $R=25\Omega$.

We know that the magnetic force acting on a segment of a conductor is given by:

F = lLB, where I is the uniform current, L is the length of the conductor and B is the magnetic field.

The maximum magnetic force that can act on the bar without breaking the circuit is:

$\begin{array}{r}{F}_B=F_g\\ =mg\\ =\left(0.750\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\\ =7.35\mathrm{N}\end{array}$

Now, putting the values in the formula we get:

$I=\frac{V}{R}=\frac{817}{2.0\mathrm{\Omega }}$

Now, putting the values in ohms law we get:

$\begin{array}{r}V=32.7\times 25\\ =817\mathrm{V}\end{array}$

Therefore, the voltage is 817 V

The current passing through the bar is:

$113\mathrm{m}/{\mathrm{s}}^2$

= 408 A

Now,

$\begin{array}{r}{\mathrm{F}}_{\mathrm{B}}=\left(408\right)(0.500)(0.4050)\\ =91.9\mathrm{N}\\ \mathrm{Applying}\mathrm{the}\mathrm{Newtons}\mathrm{second}\mathrm{law}\mathrm{we}\mathrm{get}:\\ \begin{array}{rl}\sum {\mathrm{F}}_{\mathrm{y}}& ={\mathrm{F}}_{\mathrm{B}}^{\mathrm{\prime }}-{\mathrm{F}}_{\mathrm{g}}\\ & ={\mathrm{maF}}_{\mathrm{B}}^{\mathrm{\prime }}-\mathrm{mg}\\ & ={\mathrm{F}}_{\mathrm{B}}^{\mathrm{\prime }}-{\mathrm{F}}_{\mathrm{B}}\\ & =\mathrm{ma}\\ \mathrm{a}& =\frac{{\mathrm{F}}_{\mathrm{B}}^{\mathrm{\prime }}-{\mathrm{F}}_{\mathrm{B}}}{\mathrm{m}}\end{array}\end{array}$

now, putting the values we get:

$\begin{array}{r}a=\frac{91.9N-7.35N}{0.750}\\ =113\mathrm{m}/{\mathrm{s}}^2\end{array}$

Therefore, the acceleration is$113m/s^2$