A charged particle of charge 4μC is moving perpendicular to a magnetic field with velocity 3 m/s. if it experiences a magnetic force of 3×10−3𝑁 3×10 −3 N , determine the magnitude of the magnetic field.
Question
A charged particle of charge 4μC is moving perpendicular to a magnetic field with velocity 3 m/s. if it experiences a magnetic force of 3×10−3𝑁 3×10 −3 N , determine the magnitude of the magnetic field.
Solution
The formula to calculate the magnetic field (B) is given by the equation:
F = qvBsinθ
where: F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field, and θ is the angle between the velocity and the magnetic field.
In this case, the particle is moving perpendicular to the magnetic field, so θ = 90°. The sine of 90° is 1, so the equation simplifies to:
F = qvB
We can rearrange the equation to solve for B:
B = F / (qv)
Substituting the given values:
B = 3×10−3 N / (4×10-6 C * 3 m/s)
B = 0.25 T
So, the magnitude of the magnetic field is 0.25 Tesla.
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