Q79 / 100 · BITSAT 2024 (Memory Based Paper 3) · Physics
Inside a solenoid of radius 0.5 m , magnetic field is changing at a rate of $50 \times 10^{-6} \mathrm{~T} / \mathrm{s}$.
Acceleration of an electron placed at a distance of 0.3 m from axis of solenoid will be
- Correct answerA$23 \times 10^6 \mathrm{~ms}^{-2}$
- B$26 \times 10^6 \mathrm{~m} / \mathrm{s}^2$
- C$1.3 \times 10^9 \mathrm{~ms}^{-2}$
- D$26 \times 10^9 \mathrm{~m} / \mathrm{s}^2$
Solution
The changing magnetic field induce an electric field which will accelerate the electron.
The acceleration of electron will be
$a=\frac{e E}{m} \quad[\because F=q E]$
To find electric field $E$, we will use Faraday's electromagnetic induction.
$\mathrm{emf}, \varepsilon=\frac{-d \phi}{d t}(B \cdot A)$
where, $B$ is the magnetic field and $A$ is the area of cross-section.
$\varepsilon=-A \cdot \frac{d B}{d t}=-\pi r^2 \frac{d B}{d t}$
Now, electric field, $E=-\frac{\mathrm{emf}}{\text { distance }}=-\frac{\dot{\varepsilon}}{d}$
$=\frac{\pi r^2}{d} \frac{d B}{d t}$
Now, acceleration, $a=\frac{e}{m} \frac{\pi r^2}{d} \frac{d B}{d t}$
$=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \times \frac{22}{7}$
$\begin{aligned} & \times \frac{(0.5)^2}{(0.3)} \times 50 \times 10^{-6} \\ & =23 \times 10^6 \mathrm{~ms}^{-2}\end{aligned}$
Hence, the correct answer is option: 'A'.