91Ó°ÊÓ

Given,

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

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:

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

Now, putting the values in ohms law we get:

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

Therefore, the voltage is 817 V

The current passing through the bar is:

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

= 408 A

Now,

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

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$113m/s^2$