# How would you determine the magnitude of the magnetic field and the speed of q_2?

## A charge q_1 = 20.0 µC moves with a speed of 4.00 ✕ 10^3 $m$/$s$ perpendicular to a uniform magnetic field. The charge experiences a magnetic force of 7.46 ✕ 10^-3 N. A second charge q_2 = 7.00 µC travels at an angle of 40.0° with respect to the same magnetic field and experiences a 1.90 ✕ 10^-3 N force.

Jun 7, 2016

$B = 0.09325 T$
$v = 4.5 \times {10}^{3} m {s}^{-} 1$

#### Explanation:

The force, $F$ on a charged particle, $q$, moving perpendicular to a magnetic field, $B$ is given by:
$F = B q v$
Where v= velocity of the particle.

So for ${q}_{1}$ we have:
7.46times 10^-3=B*(20times 10^-6)*(4times10^3)
$B = 0.09325 T$

If moving at an angle, $\theta$ with respect to the field then the equation needs to be modified as:
$F = B q v \sin \theta$

So for ${q}_{2}$

$1.9 \times {10}^{-} 3 = 0.09325 \cdot \left(7 \times {10}^{-} 6\right) . v . \sin 40$
$v = 4528 m {s}^{-} 1 = 4.5 \times {10}^{3} m {s}^{-} 1$