An electron is moving at a velocity of \( 3.2 \times 10^{7}\,\mathrm{m/s} \) in a magnetic field along a circular path of radius 20 cm. The magnitude of the magnetic field applied is (approximately)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 9 \times 10^{-4}\,\mathrm{T} \)
The magnetic force \( evB \) provides the centripetal force for the circular motion.
\( \displaystyle evB = \frac{mv^{2}}{r} \;\Rightarrow\; B = \frac{mv}{er} \)
\( m = 9.1\times10^{-31}\,\mathrm{kg}, \quad e = 1.6\times10^{-19}\,\mathrm{C} \)
\( r = 20\,\mathrm{cm} = 0.2\,\mathrm{m} \)
\( \displaystyle B = \frac{9.1\times10^{-31} \times 3.2\times10^{7}}{1.6\times10^{-19} \times 0.2} = \frac{2.912\times10^{-23}}{3.2\times10^{-20}} \)
\( \displaystyle B = 9.1\times10^{-4}\,\mathrm{T} \approx 9\times10^{-4}\,\mathrm{T} \)