Question

An electron in the beam of a cathode-ray tube is accelerated by a potential difference of 2.00Kv Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius What is the magnitude of the field?

Short answer

The magnitude of the field is$8.384\times {10}^{-4}\mathrm{T}$ .

Step-by-step solution

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

Determine the magnitude of the field

The formula of radius for path$\mathrm{R}=\frac{\mathrm{mv}}{\left|\mathrm{q}\right|\mathrm{B}}$by the kinetic energy formula$\mathrm{v}=\sqrt{\frac{2\mathrm{eV}}{\mathrm{m}}}$and q = e , the radius formula is

$$\begin{gathered} \mathrm{R}=\frac{\mathrm{m}}{\mathrm{eB}}\sqrt{\frac{2\mathrm{eV}}{\mathrm{m}}} \\ \mathrm{R}=\frac{1}{\mathrm{B}}\sqrt{\frac{2\mathrm{eV}}{\mathrm{m}}} \\ \mathrm{B}=\sqrt{\frac{2\mathrm{mV}}{{\mathrm{eR}}^2}} \end{gathered}$$

Put all values

$$\begin{gathered} \mathrm{B}=\sqrt{\frac{2\left(9.11\times 10-31\mathrm{kg}\right)\left(2.00\times {10}^3\mathrm{V}\right)}{\left(1.60\times {10}^{-19}\mathrm{kg}\right){\left(0.180\mathrm{m}\right)}^2}} \\ \mathrm{B}=8.384\times {10}^{-4}\mathrm{T} \end{gathered}$$

Hence, the magnitude of the field is$8.384\times {10}^{-4}\mathrm{T}$ .