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erastova [34]
2 years ago
5

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 b

eam travels at the same speed, but at a 30° angle with the magnetic field. What force is on these electrons?
A.(4.9 × 10-14 newtons) · tan(30°)
B.(4.9 × 10-14 newtons) · sin(30°)
C.(4.9 × 10-14 newtons) · cos(30°)
D.(4.9 × 10-14 newtons) · arctan(30°)
E.(4.9 × 10-14 newtons) · arccos(30°)
Physics
2 answers:
larisa86 [58]2 years ago
4 0

Answer:

B.(4.9 × 10-14 newtons) · sin(30°)

Explanation:

The magnetic force exerted on charged particles by a magnetic field is given by

F=qvB sin \theta

where

q is the charge

v is the speed of the charge

B is the magnetic field intensity

\theta is the angle between the direction of v and B

The first beam moves at right angle to the magnetic field, so \theta=90^{\circ} and the force on this beam is simply

F=qvB=4.9\cdot 10^{-14} N (1)

The second beam moves at angle of \theta=30^{\circ}. The electrons are travelling at same speed v, and the magnetic field is still the same (and the charge q is also the same, since they are electrons as well), so the magnetic force in this case is

F=qvB sin 30^{\circ} (2)

But from the previous equation we know that

qvB = 4.9\cdot 10^{-14} N

so, if we substitute into eq. (2), we find

F=(4.9\cdot 10^{-14} N) \cdot sin 30^{\circ}

____ [38]2 years ago
3 0
Force on a particle with charge q moving with velocity v at an angle θ to a magnetic field B is F=qvBsin(θ).  So B is correct
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