An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1×10−4 Wbm−2. The frequency of revolution of the electron will be(Take mass of electron=9.0×10−31kg)

f=1T=eB2πm =1.6×10−19×10−42π×9×10−31=2.8×106 Hz

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1×10−4 Wbm−2. The frequency of revolution of the electron will be(Take mass of electron=9.0×10−31kg)

f=1T=eB2πm =1.6×10−19×10−42π×9×10−31=2.8×106 Hz

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An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of $1 \times 10^{- 4} {Wbm}^{- 2}$ . The frequency of revolution of the electron will be
(Take mass of electron $= 9 .0 \times 10^{- 31} kg$ )

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$2.8 \times 10^{6} Hz$

$1.8 \times 10^{6} Hz$

$5 .6 \times 10^{5} Hz$

$1 .6 \times 10^{5} Hz$

answer is C.

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Detailed Solution

$\begin{array}{l} f = \frac{1}{T} = \frac{eB}{2 πm} \\ = \frac{1 .6 \times 10^{- 19} \times 10^{- 4}}{2 π \times 9 \times 10^{- 31}} = 2 .8 \times 10^{6} Hz \end{array}$

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An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1×10−4 Wbm−2. The frequency of revolution of the electron will be(Take mass of electron=9.0×10−31kg)

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An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of $1 \times 10^{- 4} {Wbm}^{- 2}$ . The frequency of revolution of the electron will be
(Take mass of electron $= 9 .0 \times 10^{- 31} kg$ )