Answer to Question #162761 in General Chemistry for Mazwi Casbeth

Question

A proton, travelling with a velocity 6.0 x 1 0 m/s due east, experiences a magnetic force that has a maximum magnitude of 8.0 x 10 N and a direction of due south.

a)  What are the magnitude and direction of the magnetic field causing the force?

b)  Repeat part ( a) assuming the proton is replaced by an electron

Answer 1

Solution:

Since the maximum magnitude of the force is found, assume the B field must be perpendicular to velocity as $sin(\alpha)$ is a maximum then. So the magnitude of the field can be found from the magnetic force equation.

$F_b=qvBsin(\alpha)$

Solve for B:

$B=\dfrac{F_b}{qvsin(\alpha)}=\dfrac{F_b}{ev}=\dfrac{8.0*10^{-14}}{(1.6*10^{-19}C)(6.0*10^{6}(\frac{m}{s})}$

$B=0.0833T$

The direction of the force is the result of the right-hand rule. In the figure below which is drawn from above looking down,

$F_B=qv\;x\;B$, so putting fingers along (to the right as drawn) and curing fingers through the smallest angle for up (out of the page as drawn) your right thumb will be pointing towards the bottom of the page which is south as drawn. So we know must be out of the page or up

The figure below repeats the process but this time we are looking horizontally and we still see to get a force pointing south must point up!

For electron:

The only change is the charge is now e instead of +e so the minus sign changes the direction from up to down to continue to have force in the south direction. The magnitude remains 0.0833 T