22P3Q08

8ai)

Electric field strength at a point is the force per unit positive charge acting on a small test charge

placed at that point.

8aii)

If not vacuum, then the electrons would suffer collision with gas particles.

Not all with arrive at the anode, and they also will not arrive at the designed speed.

8aiii)

\displaystyle \begin{aligned}   \text{loss in EPE}&=\text{gain in KE} \\  (1.60\times {{10}^{-19}})\Delta V&=2.48\times {{10}^{-16}} \\  \Delta V&=1550\text{ V} \end{aligned}

COMMENT: This answer must be given to 3 s.f., since 2.48 and 1.60 are both 3 s.f. numbers.

8aiv)

The electrons are emitted from the hot anode with a range of speeds.

COMMENT: This is kind of similar to how photoelectrons are emitted with a range of KE.

8bi)

\displaystyle \begin{aligned}   B&=\frac{\mu NI}{2r} \\  &=\frac{(4\pi \times {{10}^{-7}})(120)(3.5)}{0.30} \\  &=1.76\times {{10}^{-3}}\text{ T} \end{aligned}

8bii)

8ci)

The electron is moving perpendicularly to the direction of the magnetic field.

This results in a magnetic force that always acts in the direction perpendicular to the velocity of the electron.

This provides the required centripetal force for circular motion.

8cii)

\displaystyle \begin{aligned}   ({{F}_{net}}&=ma) \\  1.43Bqv&=\frac{m{{v}^{2}}}{r} \\  1.43Bqr&=mv \\  (1.43\times 1.76\times {{10}^{-7}})(1.60\times {{10}^{-19}})r&=(9.11\times {{10}^{-31}})(2.40\times {{10}^{7}}) \\  r&=0.0543\text{ m} \end{aligned}

8d)

The component of velocity parallel to the magnetic field keeps the electron moving forward at a constant speed.

The component of velocity perpendicular to the magnetic field results in circular motion in a plane perpendicular to the B-field.

The superposition of the forward motion and the circular motion results in the helical path.

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