# Question #1966c

May 11, 2017

$B = 9.8 \cdot {10}^{-} 2$ $T$

#### Explanation:

A basic force diagram shows that the two forces of concern acting on the copper rod are the force of gravity and the magnetic force. The force of the magnetic field must be in the opposite direction as the force of gravity in order for the rod to float, and the magnitudes of the two forces must be equal.

So we can see that a statement of the net force would follow:

${F}_{n e t} = \sum F = {F}_{B} - {F}_{G} = 0$

Note that ${F}_{B}$ is the magnetic force, and the net force is zero as the rod is assumed to float stationary in the field and therefore does not accelerate (Newton's second law).

We know that the force of gravity acting on a stationary object is given by ${F}_{G} = m g$, where $m$ is the mass of the object and $g$ is the gravitational acceleration constant, equal to $9.8 \frac{m}{s} ^ 2$.

The magnetic force acting on a current carrying wire is given by $F = I L B \sin \left(\theta\right)$, where $I$ is the current through the wire, $L$ is the length of the wire, $B$ is the strength of the magnetic field it is subjected to, and $\theta$ is the angle between the current direction and the magnetic field.

Therefore, we have:

$I L B \sin \left(\theta\right) = m g$

We want to find the magnetic field strength, so we can solve for B:

$B = \frac{m g}{I L \sin \left(\theta\right)}$

Because the direction magnetic field and current are perpendicular, $\theta = {90}^{o}$ and $\sin \left({90}^{o}\right) = 1$.

$\implies B = \frac{m g}{I L}$

We are given that $m = 48.0 g = 48.0 \times {10}^{-} 3 k g , I = 0.354 A$, and $L = 13.9 m$. We can calculate B:

$B = \frac{48 \cdot {10}^{-} 3 k g \cdot 9.8 \frac{m}{s} ^ 2}{0.345 A \cdot 13.9 m}$

$B = 0.098 T$

$\implies B = 9.8 \cdot {10}^{-} 2$ $T$