Question

An electron moving along the x-axis with an initial energy of 2.867 KeV enters a region of magnetic field $\overrightarrow{B}=3.01 mT \hat{k}$ at $S$ (See figure). The field extends between $x=0$ and $x=3$ cm. The electron is detected at the point $Q$ on a screen placed 8 cm away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is (Electron's charge = $1.6 \times 10^{-19}$ C, mass of electron = $9.1 \times 10^{-31}$ kg) ($\sqrt{834.87} \approx 28.896$)

Answer 1

3.69 cm

Answer 2

The problem describes the motion of an electron in a uniform magnetic field and then in a field-free region. We need to find the total vertical displacement of the electron on a screen.

1. Calculate the Kinetic Energy (KE) in Joules: Given initial energy KE = 2.867 KeV. $KE = 2.867 \times 10^3 \text{ eV}$ To convert eV to Joules, multiply by the charge of an electron ($1.6 \times 10^{-19}$ C): $KE = 2.867 \times 10^3 \times 1.6 \times 10^{-19} \text{ J}$ $KE = 4.5872 \times 10^{-16} \text{ J}$

2. Calculate the radius (R) of the circular path: The kinetic energy of the electron is related to its velocity by $KE = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2KE}{m}}$. When a charged particle moves perpendicular to a uniform magnetic field, it follows a circular path with radius $R = \frac{mv}{qB}$. Substituting $v$, we get $R = \frac{\sqrt{2m(KE)}}{qB}$. Given: Mass of electron ($m$) = $9.1 \times 10^{-31}$ kg Charge of electron ($q$) = $1.6 \times 10^{-19}$ C Magnetic field strength ($B$) = $3.01 \text{ mT} = 3.01 \times 10^{-3}$ T

First, calculate $\sqrt{2m(KE)}$: $2m(KE) = 2 \times (9.1 \times 10^{-31} \text{ kg}) \times (4.5872 \times 10^{-16} \text{ J})$ $2m(KE) = 8.348704 \times 10^{-46} \text{ kg m}^2/\text{s}^2$ $\sqrt{2m(KE)} = \sqrt{834.8704 \times 10^{-48}} \text{ kg m/s}$ Using the given approximation $\sqrt{834.87} \approx 28.896$: $\sqrt{2m(KE)} \approx 28.896 \times 10^{-24} \text{ kg m/s}$

Now, calculate $qB$: $qB = (1.6 \times 10^{-19} \text{ C}) \times (3.01 \times 10^{-3} \text{ T})$ $qB = 4.816 \times 10^{-22} \text{ C T}$

Now, calculate R: $R = \frac{28.896 \times 10^{-24} \text{ kg m/s}}{4.816 \times 10^{-22} \text{ C T}}$ $R = \frac{28.896}{4.816} \times 10^{-2} \text{ m}$ $R = 6.000 \times 10^{-2} \text{ m} = 6.00 \text{ cm}$

3. Determine the angle of deflection (θ) inside the magnetic field: The electron enters the magnetic field at $x=0$ and exits at $x=3$ cm. Let this width be $W = 3$ cm. From the geometry of the circular path, the horizontal distance covered is $W = R \sin \theta$. $\sin \theta = \frac{W}{R} = \frac{3 \text{ cm}}{6 \text{ cm}} = 0.5$ Therefore, $\theta = 30^\circ$.

4. Calculate the vertical displacement within the magnetic field ($y_{exit}$): The vertical displacement of the electron as it travels through the magnetic field (from $x=0$ to $x=W$) is given by $y_{exit} = R(1 - \cos \theta)$. $y_{exit} = 6 \text{ cm} (1 - \cos 30^\circ)$ $y_{exit} = 6 \text{ cm} (1 - \frac{\sqrt{3}}{2})$ $y_{exit} = 6 \text{ cm} (1 - 0.8660)$ $y_{exit} = 6 \text{ cm} (0.1340) = 0.804 \text{ cm}$

5. Calculate the vertical displacement after exiting the magnetic field ($y_{straight}$): After exiting the magnetic field, the electron moves in a straight line at an angle $\theta$ with the x-axis. The screen is placed 8 cm away from point S. The electron exits the field at $x=3$ cm. The remaining horizontal distance to the screen is $L - W = 8 \text{ cm} - 3 \text{ cm} = 5 \text{ cm}$. The additional vertical displacement on the screen due to this straight-line motion is $y_{straight} = (L - W) \tan \theta$. $y_{straight} = 5 \text{ cm} \times \tan 30^\circ$ $y_{straight} = 5 \text{ cm} \times \frac{1}{\sqrt{3}}$ $y_{straight} = 5 \text{ cm} \times 0.57735 = 2.88675 \text{ cm}$

6. Calculate the total distance 'd' between P and Q: The total distance 'd' on the screen is the sum of the vertical displacement within the field and the vertical displacement after exiting the field. $d = y_{exit} + y_{straight}$ $d = 0.804 \text{ cm} + 2.88675 \text{ cm}$ $d = 3.69075 \text{ cm}$

Rounding to two decimal places, $d \approx 3.69 \text{ cm}$.