91Ó°ÊÓ

when a wire carrying a current and moves inside a magnetic field, the magnetic field exerted a magnetic force on the wire. For a wire with length $l$, the exerted magnetic force is given by equation (29.26) in the form

${\stackrel{â‡\P Ä}{F}}_{wire}=\stackrel{â‡\P Ä}{Il}×\stackrel{â‡\P Ä}{B}=IlB\sin \mathrm{Î\pm }$

Where $\mathrm{Î\pm }$is the angle between the direction of the current and the magnetic field. A current that is parallel to the magnetic force is zero while it is maximum when the current is perpendicular to the magnetic field.

In the given circuit , the top and the bottom wires in the circuit are parallel to the magnetic field, so the exerted force both of them are zero

${\stackrel{â‡\P Ä}{F}}_{top}={\stackrel{â‡\P Ä}{F}}_{bottom}=0N$

We use Ohm's law to get this current as Ohm's law states that the current $I$flows through a resistance due to the potential difference $V$between the resistance.

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

Let us determine the current through the right wire using the values for $R=3\mathrm{Î\copyright }andV=15V$of the battery by

$I=\frac{V}{R}=\frac{15V}{3\mathrm{Î\copyright }}=5A.$

For the right wire the angle is $\mathrm{Î\pm }=90°$, so we plug values for $I,l,Band\mathrm{Î\pm }$into equation (1) to get ${\stackrel{â‡\P Ä}{F}}_{right}$by

$$\begin{gathered} {\stackrel{â‡\P Ä}{F}}_{right}=IlB\sin \mathrm{Î\pm } \\ =(5A)(0.10m)(0.050T)\sin 90° \\ =0.025N \end{gathered}$$

Now we get the net force on the circuit by

$$\begin{gathered} {\stackrel{â‡\P Ä}{F}}_{circuit}={\stackrel{â‡\P Ä}{F}}_{top}+{\stackrel{â‡\P Ä}{F}}_{right}+{\stackrel{â‡\P Ä}{F}}_{bottom} \\ =0+0.025N+0 \\ =0.025N \end{gathered}$$

The current diagram in the right wire is down because it moves from the positive terminal of the battery. if we apply the battery rule, we find that the force is exerted to the right.