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Question

A proton enters a magnetic field of flux density 1.5 T velocity of 20×102ms1 at an angle of 30 with the field. The force on the position is [ep=1.6×1019C]

A
2.4×1010N
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B
2.403×1012N
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C
3×105N
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D
3×104N
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Solution

The correct option is B 2.403×1012N
The magnitude of the magnetic force on a charge particle F=|q|vBsinθ
Here, q=charge on a proton =1.602×1019C
V=2×107m/sθ=300B=1.5T
Substituting all these values in the expression for F
We get,
F=1.602×1019C×2×107m/s×1.5Tsin300=2.403×1012N
Option B is correct.

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