Question

1. an electron accelerates from rest through an electric field and into a magnetic field as shown in the diagram below. the plates have a potential difference of 85 v, and the magnetic field has a magnitude of 2.50 t.

c) calculate the magnitude and direction of the magnetic force as the electrons enters the magnetic field.
d) determine the radius of the circular path that is formed by the electron in the magnetic field.

Explanation:

Part (c)

Step 1: Find the speed of the electron after accelerating through the electric field

The kinetic energy gained by the electron is equal to the electric potential energy lost: \( e\Delta V=\frac{1}{2}mv^{2} \). Solving for \( v \):
\( v = \sqrt{\frac{2e\Delta V}{m}} \), where \( e = 1.6\times10^{-19}\, \text{C} \), \( \Delta V = 85\, \text{V} \), \( m = 9.11\times10^{-31}\, \text{kg} \).
\( v=\sqrt{\frac{2\times1.6\times10^{-19}\times85}{9.11\times10^{-31}}}\approx\sqrt{\frac{2.72\times10^{-17}}{9.11\times10^{-31}}}\approx\sqrt{2.986\times10^{13}}\approx5.464\times10^{6}\, \text{m/s} \).

Step 2: Calculate the magnetic force magnitude

The magnetic force is \( F = qvB \), where \( q = e = 1.6\times10^{-19}\, \text{C} \), \( v = 5.464\times10^{6}\, \text{m/s} \), \( B = 2.50\, \text{T} \).
\( F = 1.6\times10^{-19}\times5.464\times10^{6}\times2.50 \).
\( F = 1.6\times5.464\times2.50\times10^{-13} \approx 21.856\times10^{-13} = 2.19\times10^{-12}\, \text{N} \).

Step 3: Determine the direction of the magnetic force

Using the right - hand rule: for a negative charge (electron), the velocity is to the left, and the magnetic field is out of the page (dots). The right - hand rule for positive charges: \( \vec{F}=q\vec{v}\times\vec{B} \). For negative \( q \), the force direction is opposite. Point your right hand fingers in the direction of \( \vec{v} \) (left) and curl them towards \( \vec{B} \) (out of page), the thumb points down. So the magnetic force on the electron is downward.

Part (d)

Step 1: Recall the centripetal force formula for circular motion

In the magnetic field, the magnetic force provides the centripetal force: \( F = \frac{mv^{2}}{r} \), and we also know \( F = qvB \). So \( qvB=\frac{mv^{2}}{r} \), solving for \( r \) gives \( r=\frac{mv}{qB} \).

Step 2: Substitute the values

We know \( m = 9.11\times10^{-31}\, \text{kg} \), \( v = 5.464\times10^{6}\, \text{m/s} \), \( q = 1.6\times10^{-19}\, \text{C} \), \( B = 2.50\, \text{T} \).
\( r=\frac{9.11\times10^{-31}\times5.464\times10^{6}}{1.6\times10^{-19}\times2.50} \).
First, calculate the numerator: \( 9.11\times5.464\times10^{-31 + 6}=9.11\times5.464\times10^{-25}\approx49.777\times10^{-25}=4.9777\times10^{-24} \).
Then, calculate the denominator: \( 1.6\times2.50\times10^{-19}=4\times10^{-19} \).
\( r=\frac{4.9777\times10^{-24}}{4\times10^{-19}} = 1.244\times10^{-5}\, \text{m}=1.24\times10^{-5}\, \text{m} \).

Part (c) Answer:

The magnitude of the magnetic force is \( \boldsymbol{2.19\times10^{-12}\, \text{N}} \) and the direction is downward.

Part (d) Answer:

The radius of the circular path is \( \boldsymbol{1.24\times10^{-5}\, \text{m}} \)

Answer:

Step 1: Recall the centripetal force formula for circular motion

In the magnetic field, the magnetic force provides the centripetal force: \( F = \frac{mv^{2}}{r} \), and we also know \( F = qvB \). So \( qvB=\frac{mv^{2}}{r} \), solving for \( r \) gives \( r=\frac{mv}{qB} \).

Step 2: Substitute the values

We know \( m = 9.11\times10^{-31}\, \text{kg} \), \( v = 5.464\times10^{6}\, \text{m/s} \), \( q = 1.6\times10^{-19}\, \text{C} \), \( B = 2.50\, \text{T} \).
\( r=\frac{9.11\times10^{-31}\times5.464\times10^{6}}{1.6\times10^{-19}\times2.50} \).
First, calculate the numerator: \( 9.11\times5.464\times10^{-31 + 6}=9.11\times5.464\times10^{-25}\approx49.777\times10^{-25}=4.9777\times10^{-24} \).
Then, calculate the denominator: \( 1.6\times2.50\times10^{-19}=4\times10^{-19} \).
\( r=\frac{4.9777\times10^{-24}}{4\times10^{-19}} = 1.244\times10^{-5}\, \text{m}=1.24\times10^{-5}\, \text{m} \).

Part (c) Answer:

The magnitude of the magnetic force is \( \boldsymbol{2.19\times10^{-12}\, \text{N}} \) and the direction is downward.

Part (d) Answer:

The radius of the circular path is \( \boldsymbol{1.24\times10^{-5}\, \text{m}} \)