Question

Inside a solenoid of radius \( 0.5 \) m, the magnetic field is changing at a rate of \( 50 \times 10^{-6} \) T/s. The acceleration of an electron placed at a distance of \( 0.3 \) m from the axis of the solenoid will be:

• A. \( 23 \times 10^6 \) m/s\(^2\)
• B. \( 26 \times 10^6 \) m/s\(^2\)
• C. \( 1.3 \times 10^9 \) m/s\(^2\)
• D. \( 26 \times 10^9 \) m/s\(^2\)

Hint:

The changing magnetic field inside a solenoid induces an electric field, which exerts a force on charged particles.

Answer 1

The Correct Option is A

Step 1: {Application of Faraday's Law}
\[{Induced emf} = \frac{-d\Phi}{dt} = B \cdot A\]Given \( A = \pi r^2 \), the equation becomes:\[\varepsilon = -\pi r^2 \frac{dB}{dt}\]Step 2: {Determination of Electric Field}
\[E = \frac{\varepsilon}{d} = \frac{\pi r^2}{d} \frac{dB}{dt}\]Step 3: {Calculation of Acceleration}
\[a = \frac{eE}{m}\]Upon substitution of values:\[a = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \times \frac{\pi (0.5)^2}{0.3} \times 50 \times 10^{-6}\]\[= 23 \times 10^6 { m/s}^2\]Therefore, the resulting acceleration is \( 23 \times 10^6 \) m/s\(^2\).