The boundary shown is that of a uniform magnetic field directed in the positive z direction.An electron enters the magnetic field with a velocity pointing along the x axis and exits 0.63 $\mu$s later at point A.What is the magnitude of the magnetic field?
A)18 $\mu$T
B)14 $\mu$T
C)28$\mu$T
D)34 $\mu$T
E)227 $\mu$T

Question

The boundary shown is that of a uniform magnetic field directed in the positive z direction.An electron enters the magnetic field with a velocity pointing along the x axis and exits 0.63 $\mu$s later at point A.What is the magnitude of the magnetic field? 
A)18 $\mu$T 
B)14 $\mu$T 
C)28$\mu$T 
D)34 $\mu$T 
E)227 $\mu$T

Quizplus · Accepted Answer

We can use the formula for the magnetic force on a charged particle:

F = q*v*B*sin(theta)

Where q is the charge of the particle (in this case, the charge of an electron), v is the velocity of the electron, B is the magnitude of the magnetic field, and theta is the angle between the velocity and the magnetic field. Since the magnetic field is directed in the positive z direction and the velocity is directed along the x axis, we have theta = 90 degrees.

The magnetic force causes the electron to move in a circular path, so we can use the formula for centripetal force to find the magnitude of the magnetic field:

F = ma = (mv^2)/r

Where m is the mass of the electron, v is its speed, and r is the radius of the circular path. We can solve for r in terms of v and B:

r = mv/(q*B)

Given that the electron exits the magnetic field 0.63 seconds later, we can use the distance formula to find the radius of the circle:

r = vt

Where t is the time spent in the magnetic field. Substituting the expression for r, we have:

mv/(q*B) = v*t

Solving for B, we get:

B = mv/(q*t)

Substituting the values, we get:

B = (9.1e-31 kg) * (2.0e7 m/s) / (1.6e-19 C * 0.63 s) = 1.4e-5 T = 14 µT

Therefore, the correct answer is B.

The Boundary Shown Is That of a Uniform Magnetic Field μ\muμ

The Boundary Shown Is That of a Uniform Magnetic Field $\mu$