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 μs later at point A. What is the magnitude of the magnetic field?
A)18 μT
B)14 μT
C)28 μT
D)34 μT
E)227 μT

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 μs later at point A. What is the magnitude of the magnetic field? 
A)18 μT 
B)14 μT 
C)28 μT 
D)34 μT 
E)227 μT

Quizplus · Accepted Answer

Using the formula for the radius of a charged particle in a magnetic field, we can find the magnetic field strength:
$$r=\frac{mv}{qB}$$
where $m$ is the mass of the particle, $v$ is its velocity, $q$ is its charge, and $B$ is the magnetic field strength. Solving for $B$, we get:
$$B=\frac{mv}{qr}$$
We know that the electron travels in a straight line, so its path must be a circle with radius $r$. Therefore, we can use the distance it travels in the magnetic field to find $r$:
$$r=\frac{d}{	heta}$$
where $d$ is the distance traveled and $	heta$ is the angle through which the electron is deflected. Since the electron enters and exits the magnetic field along a line perpendicular to the magnetic field, the angle through which it is deflected is 90 degrees. Therefore:
$$r=\frac{d}{90^\circ}=\frac{d}{\pi/2}=\frac{2d}{\pi}$$
Using the given distance and time, we can find the velocity of the electron:
$$v=\frac{d}{t}=\frac{2.2	ext{ mm}}{0.63	ext{ }\mu	ext{s}}=3.492	ext{ }	ext{km/s}$$
The mass and charge of an electron are $9.11	imes10^{-31}$ kg and $-1.602	imes10^{-19}$ C, respectively. Putting it all together:
$$B=\frac{(9.11	imes10^{-31}	ext{ kg})(3.492	imes10^3	ext{ m/s})}{(-1.602	imes10^{-19}	ext{ C})(2(7.5	imes10^{-3}	ext{ m}/\pi))}=1.42	imes10^{-5}	ext{ T}=14	ext{ }\mu	ext{T}$$
Therefore, the answer is choice B.

The Boundary Shown Is That of a Uniform Magnetic Field