An electron experiences a magnetic force of magnitude 4.60 x 10-15 N when moving at an angleof 600 with respect to a magnetic field of magnitude 3.50 x 10-3 T. Find the speed of theelectron.
Question
An electron experiences a magnetic force of magnitude 4.60 x 10-15 N when moving at an angleof 600 with respect to a magnetic field of magnitude 3.50 x 10-3 T. Find the speed of theelectron.
Solution 1
The force experienced by a charged particle moving in a magnetic field is given by the equation:
F = qvBsinθ
where: F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
We can rearrange this equation to solve for v:
v = F / (qBsinθ)
Given that the force F is 4.60 x 10^-15 N, the magnetic field strength B is 3.50 x 10^-3 T, the angle θ is 60 degrees, and the charge of an electron q is 1.60 x 10^-19 C, we can substitute these values into the equation:
v = (4.60 x 10^-15 N) / [(1.60 x 10^-19 C) * (3.50 x 10^-3 T) * sin(60)]
Now, calculate the value of v.
Solution 2
The force experienced by a charged particle moving in a magnetic field is given by the equation:
F = qvBsinθ
where: F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
We can rearrange this equation to solve for v:
v = F / (qBsinθ)
Given that the force F is 4.60 x 10^-15 N, the magnetic field strength B is 3.50 x 10^-3 T, the angle θ is 60 degrees, and the charge of an electron q is 1.60 x 10^-19 C, we can substitute these values into the equation:
v = (4.60 x 10^-15 N) / [(1.60 x 10^-19 C) * (3.50 x 10^-3 T) * sin(60)]
Now, calculate the value of v.
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