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Moving Charges and Magnetism — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsMoving Charges and Magnetism4 Marks
Question
An electron emitted by a heated cathode and accelerated through a potential difference of $2.0 kV,$ enters a region with uniform magnetic field of $0.15 T$. Determine the trajectory of the electron if the field,
  1. is transverse to its initial velocity,
  2. makes an angle of $30^\circ$ with the initial velocity.
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✓

Answer

Magnetic field strength $, B = 0.15 T$ Charge on the electron $, e = 1.6 \times 10^{-19} C$
Mass of the electron $, m = 9.1 \times 10^{-31} \ kg$
Potential difference $, V = 2.0 kV = 2 \times 10^3 V$
Thus, kinetic energy of the electron $= eV$
$\Rightarrow\text{eV}=\frac{1}{2}\text{mv}^{2}$
$\text{v}=\sqrt\frac{2\text{ev}}{\text{m}}.........(1)$
Where,
$v = $ velocity of the electron
  1. Magnetic force on the electron provides the required centripetal force of the electron.
  2. Hence, the electron traces a circular path of radius $r$.
Magnetic force on the electron is given by the relation,
$\text{B ev}$
$\text{Centripetal force}=\frac{\text{mv}^{2}}{\text{r}}$
$\therefore\text{Bev}=\frac{\text{mv}^{2}}{\text{r}}$
$\text{r}=\frac{\text{mv}}{\text{Be}}.........(2)$
From equations $(1)$ and $(2), $ we get
$\text{r}=\frac{\text{m}}{\text{Be}}\Big[\frac{2\text{ev}}{\text{m}}\Big]^{\frac{1}{2}}$
$=\frac{9.1\times10^{-31}}{0.15\times1.6\times10^{-19}}\times\Big(\frac{2\times1.6\times10^{-19}\times2\times10^{3}}{9.1\times10^{-31}}\Big)^{\frac{1}{2}}$
$=100.55\times10^{-5}$
$=1.01\times10^{-3}\text{ m}$
$=1\text{ mm}$
Hence, the electron has a circular trajectory of radius $1.0 \ mm$ normal to the magnetic field.
  1. When the field makes an angle $\theta$ of $30^\circ$ with initial velocity, the initial velocity will be,
$\text{v}_{1}=\text{v}\sin\theta$
From equation $(2),$ we can write the expression for new radius as: $\text{r}_{1}=\frac{\text{mv}_{1}}{\text{Be}}$
$=\frac{\text{mv}\sin\theta}{\text{Be}}$
$=\frac{9.1\times10^{-31}}{0.15\times1.6\times10^{-19}}\times\Big(\frac{2\times1.6\times10^{-19}\times2\times10^{3}}{9\times10^{-31}}\Big)^{\frac{1}{2}}\times\sin30^{\circ}$
$=0.5\times10^{-31}\text{ m}$
$=0.5\text{ mm}$
Hence, the electron has a helical trajectory of radius $0.5 \ mm$ along the magnetic field direction.

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