A particle that has a charge of 7.4 μC moves with a velocity of magnitude 3 × 105 m/s along the +x axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge with the given speed could experience has a magnitude of 0.520 N. Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.

Explanation:

F = qvB

Where F equal to the magnetic force on the particle.

q = charge on the particle

v = velocity of the particle

B =Magnetic field

F = 0.520N

q = 7.4 μC = 7.4×10-⁶C

v = 3×10⁵m/s

B= ?

B = F/qv

B = 0.520/(7.4×10-⁶ ×3×10⁵)

B = 0.234T.

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

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the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this