KingCarnage3712
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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?

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°)

Respuesta :

egenriether
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
skyluke89

Answer:

B.(4.9 × 10-14 newtons) · sin(30°)

Explanation:

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

[tex]F=qvB sin \theta[/tex]

where

q is the charge

v is the speed of the charge

B is the magnetic field intensity

[tex]\theta[/tex] is the angle between the direction of v and B

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

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

The second beam moves at angle of [tex]\theta=30^{\circ}[/tex]. 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

[tex]F=qvB sin 30^{\circ}[/tex] (2)

But from the previous equation we know that

[tex]qvB = 4.9\cdot 10^{-14} N[/tex]

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

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

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