Question

An electron in a cathode-ray tube is accelerated through a potential difference of $10kV$, then passes through the $2.0-cm$wide region of uniform magnetic field in FIGURE. What field strength will deflect the electron by $10°$?

Short answer

Field strength,$B=3\mathrm{mT}$

Step-by-step solution

Find Voltage (part a)

The electron's energy is preserved while it accelerates in the electric field and subsequently moves in the magnetic field in a cyclotron motion. The conservation law of energy is used to calculate the electron's speed when potential energy is converted to kinetic energy.

$U=E$

$e\Delta V=\frac{1}{2}m{v}^2$

$v=\sqrt{\frac{2e\Delta V}{m}}$

Find Voltage (part b)

we plug the values for$e,\Delta V$ and $m$ to get$v$ by

$v=\sqrt{\frac{2e\Delta V}{m}}=\sqrt{\frac{2\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(10\times {10}^3\mathrm{V}\right)}{9.11\times {10}^{-31}\mathrm{kg}}}$

$=6\times {10}^7\mathrm{m}/\mathrm{s}$

Find rcyc (part a)

The cyclotron motion is defined as a particle travelling in a uniform circular motion perpendicular to the magnetic field at a constant speed. The radius of the circular motion generated by the magnetism is related to it by an equation in the form

$r_{\mathrm{cyc}}=\frac{mv}{qB}$

Where $q$is the charge of the particle, $v$is the speed of the particle, $B$is the magnetic field and $m$is the mass of the particles.

The electron begins to deflect with a radius of $r_{\mathrm{cyc}}$and moves a vertical distance of $y$downward, resulting in an angle $\theta$of from the supplied figure.

$\sin \theta =\frac{y}{r_{\mathrm{cyc}}}$

$r_{\mathrm{cyc}}=\frac{y}{\sin \theta }=\frac{2\mathrm{cm}}{\sin 10°}=11.5\mathrm{cm}$

Find Field strength

Rearrange the equation:

$B=\frac{mv}{q{r}_{\mathrm{cyc}}}$

Put the values in equation,

$=\frac{\left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(6\times {10}^7\mathrm{m}/\mathrm{s}\right)}{(0.115\mathrm{m})\left(1.6\times {10}^{-19}\mathrm{C}\right)}$

$=3\times {10}^{-3}\mathrm{T}$

$=3\mathrm{mT}$