How would you determine the magnitude of the magnetic field and the speed of $q_{2}$ $q_2$ ?

Question

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.

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Answer 1 · 2016-06-07T10:36:10

$B=0.09325T$
$v=4.5times10^3 ms^-1$

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

So for $q_1$ we have:
$7.46times 10^-3=B*(20times 10^-6)*(4times10^3$ )
$B=0.09325T$

If moving at an angle, $theta$ with respect to the field then the equation needs to be modified as:
$F=Bqvsintheta$

So for $q_2$

$1.9times 10^-3=0.09325*(7times10^-6).v.sin40$
$v=4528 ms^-1 =4.5times10^3 ms^-1$