An electron beam of a TV picture tube is accelerated through a potential difference of 2.00 kV. It then passes into a magnetic field perpendicular to its path, where it moves in a circular diameter of .360 m. What is the magnitude of the field?
I am lost. I keep getting the wrong answer! Help!
First calculate the velocity V after being accelerated by the potential difference.
(1/2)m V^2 = 2000 J/coulomb*1.602*10^-19 coulombs
Here, m is the electron mass
When travelling arouind the B field line,
e V B = m V^2/R , the centripetal force that keeps it moving in a circle.
B = (m/e)(V/R)
Sure! Let's break down the problem step by step to find the magnitude of the magnetic field.
We are given the following information:
- Potential difference (V) across which the electron beam is accelerated = 2.00 kV = 2000 V
- Diameter of the circular path (d) = 0.360 m
To find the magnitude of the magnetic field, we can use the formula for the radius (r) of the circular path of a charged particle moving through a magnetic field:
r = (m*v) / (q*B)
Where:
- m is the mass of the electron
- v is the velocity of the electron
- q is the charge of the electron
- B is the magnitude of the magnetic field
To find the velocity (v) of the electron, we can use the electrostatic potential energy (PE) converted to kinetic energy (KE) equation:
PE = KE
qV = (1/2)m*v^2
Rearranging the equation to solve for v:
v = sqrt(2*qV / m)
Now we can substitute this value of v into the radius equation to solve for B:
r = (m * sqrt(2*qV / m)) / (q * B)
Simplifying:
r = sqrt(2 * V / B)
B = (2 * V) / r^2
Now let's plug in the values to calculate the magnitude of the magnetic field:
B = (2 * 2000) / (0.360)^2
Calculating this, we get:
B ≈ 30555.55 T
Therefore, the magnitude of the magnetic field is approximately 30555.55 T.
Make sure to double-check your calculations to ensure accuracy.