An electron is moving at 3.6 $\times$ 10⁶ m/s perpendicular to a uniform magnetic field.If the radius of the motion is 18 mm,what is the magnitude of the magnetic field? A) 70 mT B) 0.95 mT C) 9.3 mT D) 1.6 $\mu$T E) 8.9 nT

Question

Quizplus · Accepted Answer

The equation for the radius of motion of a charged particle in a magnetic field is:
r = mv/qB
where r is the radius of the particle's circular path, m is its mass, v is its velocity, q is its charge, and B is the magnitude of the magnetic field.
In this problem, the electron's velocity is given as 3.6×10^6 m/s, its charge is -1.6×10^-19 C, and its radius of motion is 18 mm, or 0.018 m.
Substituting these values into the equation above, we can solve for B:
B = mv/qr = (9.109×10^-31 kg)(3.6×10^6 m/s)/(-1.6×10^-19 C)(0.018 m) ≈ 9.5×10^-4 T ≈ 0.95 mT
Therefore, the correct answer is B.

An Electron Is Moving at 3 ×\times× 106 M/s Perpendicular to a Uniform Magnetic Field

An Electron Is Moving at 3 $\times$ 10⁶ M/s Perpendicular to a Uniform Magnetic Field