Question

A particle of mass $ 2.2 \times 10^{-30}\, \text{kg} $ and charge $ 1.6 \times 10^{-19}\, \text{C} $ is moving at $ 10\, \text{km/s} $ in a circular path of radius $ 2.8\, \text{cm} $ inside a solenoid. The solenoid has $ 25\, \text{turns/cm} $, and the magnetic field is perpendicular to the plane of the particle’s path. The current in the solenoid is:
($ \mu_0 = 4\pi \times 10^{-7}\, \text{H/m} $)

• A. 1.25 mA
• B. 10.20 mA
• C. 2.50 mA
• D. 1.56 mA

Hint:

Use \( B = \frac{mv}{qr} \) for circular motion in magnetic field and \( B = \mu_0 n I \) for solenoids.

Answer 1

The Correct Option is D

Centripetal force is provided by magnetic Lorentz force: \[ \frac{mv^2}{r} = qvB \Rightarrow B = \frac{mv}{qr} \] Given: - \( m = 2.2 \times 10^{-30} \, \text{kg} \)
- \( v = 10^4 \, \text{m/s} \)
- \( q = 1.6 \times 10^{-19} \, \text{C} \)
- \( r = 2.8 \, \text{cm} = 2.8 \times 10^{-2} \, \text{m} \)
\[ B = \frac{2.2 \times 10^{-30} \cdot 10^4}{1.6 \times 10^{-19} \cdot 2.8 \times 10^{-2}} \approx 4.91 \times 10^{-5}\, \text{T} \] Magnetic field inside solenoid: \[ B = \mu_0 n I \Rightarrow I = \frac{B}{\mu_0 n} \] Convert \( n = 25\, \text{turns/cm} = 2500\, \text{turns/m} \) \[ I = \frac{4.91 \times 10^{-5}}{4\pi \times 10^{-7} \cdot 2500} \approx 1.56 \times 10^{-3} = 1.56\, \text{mA} \]