An electron in the beam of a TV picture tube is accelerated by a potential difference of `2.00 kV`. Then, it passes through region of transverse magnetic field, where it moves in a circular arc with radius `0.180 m`. at is the magnitude of the field?

Explanation

To find the magnitude of the magnetic field through which an electron moves in a circular arc after being accelerated by a potential difference, we can follow these steps:

### Step 1: Understand the relationship between potential difference and kinetic energy
When an electron is accelerated through a potential difference \( V \), it gains kinetic energy given by:
\[ KE = eV \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) C) and \( V \) is the potential difference in volts.

### Step 2: Calculate the kinetic energy
Given that the potential difference is \( 2.00 \, \text{kV} = 2000 \, \text{V} \):
\[ KE = eV = (1.6 \times 10^{-19} \, \text{C})(2000 \, \text{V}) = 3.2 \times 10^{-16} \, \text{J} \]

### Step 3: Relate kinetic energy to velocity
The kinetic energy of the electron can also be expressed in terms of its mass \( m \) and velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = 3.2 \times 10^{-16} \, \text{J} \]

### Step 4: Solve for velocity \( v \)
Rearranging the equation to solve for \( v \):
\[ v^2 = \frac{2 \times 3.2 \times 10^{-16}}{m} \]
\[ v = \sqrt{\frac{2 \times 3.2 \times 10^{-16}}{m}} \]
The mass of the electron \( m \) is approximately \( 9.1 \times 10^{-31} \, \text{kg} \):
\[ v = \sqrt{\frac{2 \times 3.2 \times 10^{-16}}{9.1 \times 10^{-31}}} \]

### Step 5: Calculate the velocity \( v \)
Calculating the value:
\[ v \approx \sqrt{\frac{6.4 \times 10^{-16}}{9.1 \times 10^{-31}}} \approx \sqrt{7.02 \times 10^{14}} \approx 8.38 \times 10^{7} \, \text{m/s} \]

### Step 6: Use the magnetic force to find the magnetic field \( B \)
The magnetic force provides the centripetal force necessary for circular motion:
\[ F = qvB = \frac{mv^2}{R} \]
where \( R = 0.180 \, \text{m} \) is the radius of the circular path. Rearranging gives:
\[ B = \frac{mv}{qR} \]

### Step 7: Substitute values into the equation
Substituting \( m \), \( v \), \( q \), and \( R \):
\[ B = \frac{(9.1 \times 10^{-31} \, \text{kg})(8.38 \times 10^{7} \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C})(0.180 \, \text{m})} \]

### Step 8: Calculate the magnetic field \( B \)
Calculating this gives:
\[ B \approx \frac{7.63 \times 10^{-23}}{2.88 \times 10^{-20}} \approx 2.65 \times 10^{-3} \, \text{T} \approx 0.00265 \, \text{T} \]

### Final Answer
The magnitude of the magnetic field is approximately:
\[ B \approx 8.38 \times 10^{-4} \, \text{T} \text{ or } 0.838 \, \text{mT} \]