Answer to Question #159805 in Electric Circuits for zain ul abdeen

Question

The picture tube in a television uses magnetic deflection coils rather than electric deflection plates. Suppose an electron beam is accelerated through a 50.0-kV potential difference and then through a region of uniform magnetic field 1.00 cm wide. The screen is located 10.0 cm from the center of the coils and is 50.0 cm wide. When the field is turned off, the electron beam hits the center of the screen. What field magnitude is necessary to deflect the beam to the side of the screen? Ignore relativistic corrections. 

Answer 1

Explanations & Calculations

• What happens here is the electron enters the magnetic region after accelerated through the electric field & then deflect due to the magnetic field.
• Within the magnetic region, it undergoes a circular motion & exit at the desired orientation so as to hit the far end of the tube. Consider the figure attached

• Desired orientation is such that it makes $\small \theta=\tan^{-1}(\frac{25cm}{10cm})$ form the center of the coil.

• Now the velocity at it enters the magnetic region can be calculated to be,

$\qquad\qquad \begin{aligned} \small q\Delta V&= \small \frac{1}{2}mv^2\\ \small v &= \small \sqrt{2(50\times10^3V)(1.76\times10^{11}Ckg^{-1})}\\ &= \small 1.33\times10^8ms^{-1} \end{aligned}$

• For the circular motion within the magnetic region apply Newton's second law,

$\qquad\qquad \begin{aligned} \small \uparrow F&= \small ma\\ \small Bqv&= \small m\frac{v^2}{r}\\ \small B &= \small \frac{mv^2}{rqv}\\ &= \small \frac{2q\Delta V}{rqv}\\ &= \small \frac{2\Delta V}{rv}\cdots(1) \end{aligned}$ substituting from the previous equation

• $\small r$ : the radius of the moving path within the magnetic region has to be found and by the geometry there as shown it is,

$\qquad\qquad \begin{aligned} \small r \sin \theta &= \small 0.01m\\ \small r&= \small \frac{0.01}{\sin \theta}\\ &= \small \frac{0.01}{0.9285}\\ &= \small 0.0108m \end{aligned}$

• Now by (1),

$\qquad\qquad \begin{aligned} \small B&= \small \frac{2(50\times10^3V)}{(0.0108m)(1.33\times10^8ms^{-1})}\\ &= \small \bold{0.0696\,T} \end{aligned}$