An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 × 10 - 2 T . If, the value of e m is 1 .76 × 10 11 C kg - 1 , the frequency of revolution of the electron isNEET 2016 (Phase 2)

f = q B 2 π m = 1.76 × 10 11 × 3.57 × 10 - 2 2 × 3.14 = 10 9 Hz = 1 GHz .

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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMagnetic ForceMedium2 minPYQ_2016

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An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{- 2} T$ . If, the value of $\frac{e}{m}$ is $1 .76 \times 10^{11} C {kg}^{- 1}$ , the frequency of revolution of the electron isNEET 2016 (Phase 2)

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Answer:

Solution:

$f = \frac{q B}{2 π m} = \frac{1.76 \times 10^{11} \times 3.57 \times 10^{- 2}}{2 \times 3.14}$

$= 10^{9} Hz = 1 GHz$ .

Stream:NEETSubject:PhysicsTopic:MagnetismSubtopic:Magnetic Force

⏱ 2mℹ️ Source: PYQ_2016

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