Question

WILL MEDAL!! One beam of electrons moves at right angles to a magnetic field. The force on these electrons is 4.9 × 10-14 newtons. A second beam travels at the same speed, but at a 30° angle with the magnetic field. What force is on these electrons?
(4.9 × 10-14 newtons) · tan(30°)
(4.9 × 10-14 newtons) · sin(30°)
(4.9 × 10-14 newtons) · cos(30°)
(4.9 × 10-14 newtons) · arctan(30°)
(4.9 × 10-14 newtons) · arccos(30°)

Answer 1

try the 4th one

Answer 2

there is two situation...
1)F=qvBsin(theta)
in this situation F=4.9*10^-14 N
theta =90 degree.....so sin(theta)=1
q(magnitude)=1.6*10^-19 C
find B

Answer 3

actually its the second one

Answer 4

sin30=0.5

Answer 5

the 4th one

Answer 6

yes

Answer 7

its the second one

Answer 8

Thanks @Yopelletsunachan1

Answer 9

4th

Answer 10

b i agree @Yopelletsunachan1

Answer 11

your welcome

Answer 12

I would do this all algebraically: \[F_1=qvB \sin \theta _1\]\[F_2=qvB \sin \theta _2\] Since q the charge of the electron, v the speed, and B the magnetic field don't change we can pretty easily divide the first equation from the second equation to see: \[\frac{F_2}{F_1}=\frac{\sin \theta_2}{\sin \theta _1}\] now just multiply F_1 by both sides and plug in theta 1 and theta 2\[F_2=\frac{\sin \theta_2}{\sin \theta _1}F_1=\frac{\sin 30}{\sin 90}(4.9*10^{-14})\]
Sort of saves a lot of pain if you can do it this way I think.