Question 4: Electromagnetic Induction

Question

Inside a solenoid of radius 0.5 m , magnetic field is changing at a rate of $ 50 	imes 10^{-6} \mathrm{~T} / \mathrm{s} $. Acceleration of an electron placed at a distance of 0.3 m from axis of solenoid will be

Accepted Answer

The changing magnetic field induces an electric field, which accelerates the electron.
To determine the acceleration of the electron, we use the formula:
$ a = \frac{eE}{m} $
where $ F = qE $.
To calculate the electric field $ E $, we apply Faraday's law of electromagnetic induction:
$ \varepsilon = -\frac{d\Phi}{dt} = -B \cdot A $
Here, $ B $ is the magnetic field, and $ A $ is the cross-sectional area. Therefore, the electromotive force (emf) is:
$ \varepsilon = -A \cdot \frac{dB}{dt} = -\pi r^2 \cdot \frac{dB}{dt} $
The electric field $ E $ is then given by:
$ E = -\frac{\varepsilon}{	ext{distance}} = -\frac{\varepsilon}{d} $
$ E = \frac{\pi r^2}{d} \cdot \frac{dB}{dt} $
Finally, the acceleration $ a $ is:
$ a = \frac{e}{m} \cdot \frac{\pi r^2}{d} \cdot \frac{dB}{dt} $
Substituting the given values:
$ \begin{aligned} a &= \frac{1.6 	imes 10^{-19}}{9.1 	imes 10^{-31}} 	imes \frac{22}{7} 	imes \frac{(0.5)^2}{0.3} 	imes 50 	imes 10^{-6} \ &= 23 	imes 10^6 \, \mathrm{ms}^{-2} \end{aligned} $

Answer

$ 23 	imes 10^{6} \mathrm{~ms}^{-2} $

Answer

$ 26 	imes 10^{6} \mathrm{~m} / \mathrm{s}^{2} $

Answer

$ 1.3 	imes 10^{9} \mathrm{~ms}^{-2} $

Answer

$ 26 	imes 10^{9} \mathrm{~m} / \mathrm{s}^{2} $

Electromagnetic Induction

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

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