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Moving Charges and Magnetism — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMoving Charges and Magnetism5 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 30º with the initial velocity.
✓

Answer

Magnetic field strength, $B = 0.15$ TCharge 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. 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 θ of 30° 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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