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
$(a)$ is transverse to its initial velocity,
$(b)$ makes an angle of $30^o$ with the initial velocity
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
$(a)$ is transverse to its initial velocity,
$(b)$ makes an angle of $30^o$ with the initial velocity
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 $=e V$
$\Rightarrow e V=\frac{1}{2} m v^{2}$
$v=\sqrt{\frac{2 e V}{m}}\dots(i)$
Where,
$v=$ Velocity of the electron
$(a)$ 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,$ Bev$
Centripetal force $=\frac{m v^{2}}{r}$
$\therefore B e v=\frac{m v^{2}}{r}$
$r=\frac{m v}{B e}\dots(ii)$
From equations $(i)$ and $(ii)$, we get
$r=\frac{m}{B e}\left[\frac{2 e V}{m}\right]^{1 / 2}$
$=\frac{9.1 \times 10^{-31}}{0.15 \times 1.6 \times 10^{-19}} \times\left(\frac{2 \times 1.6 \times 10^{-19} \times 2 \times 10^{3}}{9.1 \times 10^{-31}}\right)^{1 / 2}$
$=100.55 \times 10^{-5}$
$=1.01 \times 10^{-3} \,m$
$=1\, m,m$
Hence, the electron has a circular trajectory of radius $1.0\, m\,m$ normal to the magnetic field.
$(b)$ When the field makes an angle $\theta$ of $30^{\circ}$ with initial velocity, the initial velocity will be, $v_{1}=v \sin \theta$
From equation $(ii)$, we can write the expression for new radius as:
$r_{1}=\frac{m v_{1}}{B e}$
$=\frac{m v \sin \theta}{B e}$
$=\frac{9.1 \times 10^{-31}}{0.15 \times 1.6 \times 10^{-19}} \times\left[\frac{2 \times 1.6 \times 10^{-19} \times 2 \times 10^{3}}{9 \times 10^{-31}}\right]^{\frac{1}{2}} \times \sin 30^{\circ}$
$=0.5 \times 10^{-3}\, m$
$=0.5 \,m\,m$
Hence, the electron has a helical trajectory of radius $0.5 \,m\,m$, with axis of the solenoid along the magnetic field direction.
Similar Questions
A particle having the same charge as of electron moves in a circular path of radius $0.5
\,cm$ under the influence of a magnetic field of $0.5\,T.$ If an electric field of $100\,V/m$ makes it to move in a straight path, then the mass of the particle is (given charge of electron $= 1.6 \times 10^{-19}\, C$ )
A particle having the same charge as of electron moves in a circular path of radius $0.5
\,cm$ under the influence of a magnetic field of $0.5\,T.$ If an electric field of $100\,V/m$ makes it to move in a straight path, then the mass of the particle is (given charge of electron $= 1.6 \times 10^{-19}\, C$ )
- [JEE MAIN 2019]
A beam of protons with speed $4 \times 10^{5}\, ms ^{-1}$ enters a uniform magnetic field of $0.3\, T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to....$cm$
(Mass of the proton $=1.67 \times 10^{-27}\, kg$, charge of the proton $=1.69 \times 10^{-19}\,C$)
A beam of protons with speed $4 \times 10^{5}\, ms ^{-1}$ enters a uniform magnetic field of $0.3\, T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to....$cm$
(Mass of the proton $=1.67 \times 10^{-27}\, kg$, charge of the proton $=1.69 \times 10^{-19}\,C$)
- [JEE MAIN 2020]
Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively $r_p, r_d$ and $r_{\alpha}$ Which one of the following relation is correct?
Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively $r_p, r_d$ and $r_{\alpha}$ Which one of the following relation is correct?
- [IIT 1997]
In the product
$\overrightarrow{\mathrm{F}} =\mathrm{q}(\vec{v} \times \overrightarrow{\mathrm{B}})$
$=\mathrm{q} \vec{v} \times\left(\mathrm{B} \hat{i}+\mathrm{B} \hat{j}+\mathrm{B}_{0} \hat{k}\right)$
For $\mathrm{q}=1$ and $\vec{v}=2 \hat{i}+4 \hat{j}+6 \hat{k}$ and
$\overrightarrow{\mathrm{F}}=4 \hat{i}-20 \hat{j}+12 \hat{k}$
What will be the complete expression for $\vec{B}$ ?
In the product
$\overrightarrow{\mathrm{F}} =\mathrm{q}(\vec{v} \times \overrightarrow{\mathrm{B}})$
$=\mathrm{q} \vec{v} \times\left(\mathrm{B} \hat{i}+\mathrm{B} \hat{j}+\mathrm{B}_{0} \hat{k}\right)$
For $\mathrm{q}=1$ and $\vec{v}=2 \hat{i}+4 \hat{j}+6 \hat{k}$ and
$\overrightarrow{\mathrm{F}}=4 \hat{i}-20 \hat{j}+12 \hat{k}$
What will be the complete expression for $\vec{B}$ ?
- [NEET 2021]
An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement$(s)$ is/are true?
$(A)$ They will never come out of the magnetic field region.
$(B)$ They will come out travelling along parallel paths.
$(C)$ They will come out at the same time.
$(D)$ They will come out at different times.
An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement$(s)$ is/are true?
$(A)$ They will never come out of the magnetic field region.
$(B)$ They will come out travelling along parallel paths.
$(C)$ They will come out at the same time.
$(D)$ They will come out at different times.
- [IIT 2011]