Question

A 5.0 m portion of wire carries a current of 8.0 A from east to west. It experiences a magnetic field of 6.0×10^−4 T running from north to south. What is the magnitude and direction of the magnetic force on the wire?1.2×10^−2N downward2.4×10^−2N upward1.2×10^−2N upward2.4×10^−2N downward

Answer 1

Given data

The length of the wire is L = 5 m

The magnitude of the current is I = 8 A

The magnitude of the magnetic field is B = 6 x 10-4 T

The expression for the magnitude of the magnetic force is given as:

$F=(I\times L)\times B$

Substitute the value in the above equation.

$\begin{gathered} F=8\text{ A}\times5\text{ m}\times6\times10^{-4}\text{ T} \\ F=2.4\times10^{-2}\text{ N} \end{gathered}$

Thus, the magnitude of the magnetic force is 2.4 x 10-2 N.

The direction of the magnetic field is given as:

$\vec{B}=B(-\text{ }\hat{\text{j }}\text{)}$

The direction of the current is given as:

$\vec{I}=I(-\hat{i})$

The direction of the magnetic force is given as:

$\begin{gathered} \vec{F}=L(\vec{I}\times\vec{B}) \\ \vec{F}=L(I(-\hat{i})\times B(-\hat{j}) \\ \vec{F}=LIB(\hat{k}) \end{gathered}$

The positive direction means the direction of the force is outside the page.

Thus, the direction of the magnetic force is upward.