"An alpha particle (He2+) with a velocity of 2.6x106m/scrosses a uniform magnetic field at an angle of 37.0°to the field lines and experiences a magnetic force of 1.4x10-3N. Find the strength of the magnetic field."
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Ответ:
B = 5.59x10⁹ T
Explanation:
The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:
We have:
F = 1.4x10⁻³ N
v = 2.6x10⁶ m/s
θ = 37.0°
q = 2*p = 2*1.6x10⁻¹⁹ C
Hence, the strength of the magnetic field is:
Therefore, the strength of the magnetic field is 5.59x10⁹ T.
I hope it helps you!
Ответ:
B
Explanation: