A metal bar, 0.850m long and having a resistance of 10.0, rests horizontally on conducting wires connecting it to the circuit shown in Figure. The bar is in a uniform, horizontal, 1.60 T magnetic field and is not attached to the wires in the circuit. What is the acceleration of the bar just after the switch S is closed?

The term current may be defined as the ratio of charge and time.

The equivalent resistance can be calculated as

$\begin{array}{r}\frac{1}{{\mathrm{R}}_{\mathrm{b}2}}=\frac{1}{{\mathrm{R}}_{\mathrm{b}}}+\frac{1}{{\mathrm{R}}_2}\\ \frac{1}{{\mathrm{R}}_{\mathrm{b}2}}=\frac{1}{10.0\mathrm{\Omega }}+\frac{1}{10.0\mathrm{\Omega }}\\ {\mathrm{R}}_{\mathrm{b}2}=5.00\mathrm{\Omega }\end{array}$

The equivalent resistor of the circuit is

$\begin{array}{r}\mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_{\mathrm{b}2}\\ \mathrm{R}=25.0\mathrm{\Omega }+5.00\mathrm{\Omega }\\ \mathrm{R}=30.0\mathrm{\Omega }\end{array}$

Now using Ohm’s law

$\begin{array}{r}{\mathrm{I}}_{\text{tot}}=\frac{\mathrm{V}}{R}\\ {\mathrm{I}}_{\text{tot}}=\frac{120\mathrm{V}}{30.0\mathrm{\Omega }}\\ {\mathrm{I}}_{\text{tot}}=4.00\mathrm{A}\end{array}$

and the resistance of the bar area equal and parallel, the current${\mathbf{l}}_{\mathbf{tot}}$equally divide in all of them.

So

$\begin{array}{r}{\mathrm{I}}_{\text{tot}}=\frac{{\mathrm{I}}_{\text{tot}}}{2}\\ {\mathrm{I}}_{\text{tot}}=\frac{4.00\mathrm{A}}{2}\\ {\mathrm{I}}_{\text{tot}}=2.00\mathrm{A}\end{array}$

The magnet force on the bar is due to current flow through the magnetic bar which produce magnetic field.

From the figure we can see that current flowing through the bar is directed downward and the magnetic field is directed into the screen.

Now put all the value in result of force of magnetic field

$\begin{array}{r}{\mathrm{F}}_{\mathrm{B}}=\mathrm{IBL}\\ {\mathrm{F}}_{\mathrm{B}}=(2.00\mathrm{A})(0.850\mathrm{m})(1.60\mathrm{T})\\ {\mathrm{F}}_{\mathrm{B}}=2.72\mathrm{N}\\ \mathrm{And}\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{bar}\mathrm{is}\\ \mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}}\\ \mathrm{m}=\frac{2.60\mathrm{N}}{9.80\mathrm{m}/{\mathrm{s}}^2}\\ \mathrm{m}=0.265\mathrm{kg}\end{array}$

Now the acceleration of the bar is

$\begin{array}{r}\mathrm{a}=\frac{{\mathrm{F}}_{\mathrm{B}}}{\mathrm{m}}\\ \mathrm{a}=\frac{2.72\mathrm{N}}{0.265\mathrm{kg}}\\ \mathrm{a}=10.3\mathrm{m}/{\mathrm{s}}^2\end{array}$

Hence, the acceleration of the bar just after the switch S is closed is$10.3\mathrm{m}/{\mathrm{s}}^2$to the right.