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Question: An electron is moving in a cyclotron at a speed of \(3.2 \times 10 ^ { 7 } \mathrm {~m} \mathrm {~s}...

An electron is moving in a cyclotron at a speed of 3.2×107 m s−13.2 \times 10 ^ { 7 } \mathrm {~m} \mathrm {~s} ^ { - 1 }in a magnetic field of 5×10−45 \times 10 ^ { - 4 } T perpendicular to it. What is the frequency of this electron?

A

1.4×10−5 Hz1.4 \times 10 ^ { - 5 } \mathrm {~Hz}

B

1.4×107 Hz1.4 \times 10 ^ { 7 } \mathrm {~Hz}

C

1.4×106 Hz1.4 \times 10 ^ { 6 } \mathrm {~Hz}

D

1.4×109 Hz1.4 \times 10 ^ { 9 } \mathrm {~Hz}

Answer

1.4×107 Hz1.4 \times 10 ^ { 7 } \mathrm {~Hz}

Explanation

Solution

v=3.2×107 ms−1\mathrm { v } = 3.2 \times 10 ^ { 7 } \mathrm {~ms} ^ { - 1 }

B=5×10−4 TB = 5 \times 10 ^ { - 4 } \mathrm {~T}

The frequency of electron is

v=1.4×107 Hz=14MHzv = 1.4 \times 10 ^ { 7 } \mathrm {~Hz} = 14 \mathrm { MHz }