Question

A 2.00 m length of wire carrying 3.00 A of conventional current southward through a 5.00 T magnetic field directed straight up experiences how much due the field and in what direction?

Answer 1

The magnetic force $F$ on a wire carrying current $I$ and having a length of $l$ in a magnetic field $B$ is given by ,

$F=BIl\sin \theta$ , where $\theta =$ angle between $\overrightarrow{B}$ and $\overrightarrow{l}I$

now given $l=2.00m,I=3.00A,B=5.00T$ and $\theta ={90}^o$ (as current is southward and magnetic field is straight up)

therefore $F=5.00\times 3.00\times 2.00\sin 90=30N$

the direction of force will be given by Fleming's left hand rule , and it will be perpendicular to the right of wire length .