The picture tube in an old black-and-white television uses magnetic deflection coils rather than electric deflection plates. Suppose an electron beam is accelerated through a 53.8-kV potential difference and then through a region of uniform magnetic field 1.00 cm wide. The screen is located 10.8 cm from the center of the coils and is 51.0 cm wide. When the field is turned off, the electron beam hits the center of the screen. Ignoring relativistic corrections, what field magnitude is necessary to deflect the beam to the side of the screen?
To calculate the field magnitude necessary to deflect the beam to the side of the screen, we can use the principles of magnetic deflection.
First, let's gather the relevant information from the problem:
- Potential difference (voltage) across the electron beam: 53.8 kV
- Width of the region with a uniform magnetic field: 1.00 cm
- Distance from the center of the coils to the screen: 10.8 cm
- Width of the screen: 51.0 cm
To solve this problem, we can use the fact that the magnetic force experienced by a charged particle moving in a magnetic field is given by the formula:
F = qvB
Where:
- F is the magnetic force
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnetic field strength
In this case, the electron beam is moving horizontally towards the screen, so the force acting on it needs to be directed horizontally in the opposite direction.
The force required to deflect the electron beam to the side of the screen is provided by the electric field in the potential difference (voltage) between the deflection plates. Therefore, we can equate the electrical and magnetic forces and solve for the magnetic field strength (B):
F_electric = F_magnetic
qE = qvB
Since the charge (q) and the velocity (v) of the electron beam do not change, we can simplify the equation to:
E = vB
Now, let's calculate the velocity of the electron beam using the voltage (V) and the potential energy (U) gained by the electron beam:
V = U/q
Given that the potential difference (voltage) is 53.8 kV and the energy gained by the electron beam is equal to its charge multiplied by the voltage, we can calculate the velocity:
U = qV
Next, let's calculate the magnetic field strength (B) using the formula:
E = vB
Since the electric field (E) is equal to the voltage (V) divided by the distance (d) between the deflection plates, we can substitute the values:
V/d = vB
Finally, solving for B:
B = V/(vd)
Substituting the given values:
V = 53.8 kV = 53.8 × 10^3 V
d = 1.00 cm = 1.00 × 10^−2 m
v = 10.8 cm = 10.8 × 10^−2 m
B = (53.8 × 10^3 V) / ((1.00 × 10^−2 m)(10.8 × 10^−2 m))
Now, let's calculate the value of B.