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

Concept: When a charged particle moves perpendicular to a magnetic field, it experiences a magnetic force providing the centripetal force. \[ qvB = \frac{mv^2}{r} \] Rearranging, \[ r = \frac{mv}{qB} \] where \(m\) = mass of electron \(= 9.11 \times 10^{-31}\,\text{kg}\), \(q\) = charge of electron \(= 1.6 \times 10^{-19}\,\text{C}\).

Step 1: Substitute the given values. \[ r = \frac{(9.11 \times 10^{-31})(3.2 \times 10^7)} {(1.6 \times 10^{-19})(12 \times 10^{-4})} \]

Step 2: Simplify the expression. \[ r \approx 0.15\,\text{m} \] Thus, the radius of the circular path is approximately \(0.15\) m.