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Question: If a electron is moving in a magnetic field of \(5.4 \times 10 ^ { - 4 }\) Ton a circular path of r...

If a electron is moving in a magnetic field of 5.4×10−45.4 \times 10 ^ { - 4 } Ton a circular path of radius 32 cm having a frequency of 2.5MHz, then its speed will be

A

8.56×106 m s−18.56 \times 10 ^ { 6 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

B

5.024×106 m s−15.024 \times 10 ^ { 6 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

C

8.56×104 m s−18.56 \times 10 ^ { 4 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

D

5.024×104 m s−15.024 \times 10 ^ { 4 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

Answer

5.024×106 m s−15.024 \times 10 ^ { 6 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

Explanation

Solution

Here, B=5.4×10−4 TB = 5.4 \times 10 ^ { - 4 } \mathrm {~T}

r=32 cm=32×10−2 m\mathrm { r } = 32 \mathrm {~cm} = 32 \times 10 ^ { - 2 } \mathrm {~m}

v=2.5MHz=2.5×106 Hzv = 2.5 \mathrm { MHz } = 2.5 \times 10 ^ { 6 } \mathrm {~Hz}

The speed of electron on circular path

=32×10−2×2×3.14×2.5×106= 32 \times 10 ^ { - 2 } \times 2 \times 3.14 \times 2.5 \times 10 ^ { 6 }

=502.4×104=5.024×106 ms−1= 502.4 \times 10 ^ { 4 } = 5.024 \times 10 ^ { 6 } \mathrm {~ms} ^ { - 1 }