Question:
In an experiment, electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = \(1.6 \times 10^{-19}\) C Mass of the electron = \(9.1 \times 10^{-31}\) kg]
In an experiment, electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = \(1.6 \times 10^{-19}\) C Mass of the electron = \(9.1 \times 10^{-31}\) kg]
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Get the speed from \(eV = \frac{1}{2}mv^2\), then use \(r = \frac{mv}{eB}\).
Updated On: Nov 5, 2026
- \(7.5 \times 10^{-4}\) m
- \(7.5 \times 10^{-3}\) m
- 7.5 m
- \(7.5 \times 10^{-2}\) m
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The Correct Option is A
Solution and Explanation
Kinetic energy gained: \(eV = \frac{1}{2}mv^2\) so \(v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2\times1.6\times10^{-19}\times500}{9.1\times10^{-31}}} = \sqrt{1.758\times10^{14}} = 1.326\times10^{7}\) m/s.
In the magnetic field \(B = 100\) mT = 0.1 T: \(r = \frac{mv}{eB} = \frac{9.1\times10^{-31}\times1.326\times10^{7}}{1.6\times10^{-19}\times0.1}\).
\(r = \frac{1.207\times10^{-23}}{1.6\times10^{-20}} = 7.5\times10^{-4}\) m.
Answer: (1) \(7.5 \times 10^{-4}\) m
In the magnetic field \(B = 100\) mT = 0.1 T: \(r = \frac{mv}{eB} = \frac{9.1\times10^{-31}\times1.326\times10^{7}}{1.6\times10^{-19}\times0.1}\).
\(r = \frac{1.207\times10^{-23}}{1.6\times10^{-20}} = 7.5\times10^{-4}\) m.
Answer: (1) \(7.5 \times 10^{-4}\) m
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