Question

Calculate the radius of a circular path for an electron moving through a magnetic field \(B = 12 \times 10^{-4}\,\text{T}\) with \(v = 3.2 \times 10^{7}\,\text{m/s}\).

• A. \(0.05\) m
• B. \(0.10\) m
• C. \(0.15\) m
• D. \(0.20\) m

Hint:

In magnetic field problems, remember the circular motion relation \( r=\frac{mv}{qB} \). Higher velocity or mass increases the radius, while stronger magnetic field decreases it.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A moving charge in a perpendicular magnetic field follows a circular path. We need to find the radius of this trajectory.
Step 2: Key Formula or Approach:
The magnetic force acts as the centripetal force:
\[ qvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB} \] Constants for electron: \( m = 9.1 \times 10^{-31}\ \text{kg} \), \( q = 1.6 \times 10^{-19}\ \text{C} \).
Step 3: Detailed Explanation:
Substitute the given values into the radius formula:
\[ r = \frac{(9.1 \times 10^{-31}) \times (3.2 \times 10^7)}{(1.6 \times 10^{-19}) \times (12 \times 10^{-4})} \] Simplify the coefficients:
\[ r = \frac{9.1 \times 3.2}{1.6 \times 12} \times \frac{10^{-24}}{10^{-23}} \] \[ r = \frac{9.1 \times 2}{12} \times 10^{-1} = \frac{18.2}{12} \times 0.1 \] \[ r \approx 1.516 \times 0.1 = 0.1516\ \text{m} \] Step 4: Final Answer:
The radius is approximately \( 0.15 \) m.