A proton enters a magnetic field of flux density 1.5 T with a velocity of 20 × 107ms–1 at an angle of 300 with the field. The force on the proton is [ep = 1.6×10–19C]

F=qVBsin⁡θF=1.6×10−19×20×107×1.5×12=2.4×10−11N

A proton enters a magnetic field of flux density 1.5 T with a velocity of 20 × 107ms–1 at an angle of 300 with the field. The force on the proton is [ep = 1.6×10–19C]

F=qVBsin⁡θF=1.6×10−19×20×107×1.5×12=2.4×10−11N

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A proton enters a magnetic field of flux density 1.5 T with a velocity of 20 × 10⁷ms^–1 at an angle of 30⁰ with the field. The force on the proton is [e_p = 1.6×10^–19C]

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$2.4 \times 10^{- 11} N$

$2.4 \times 10^{- 10} N$

$3 \times 10^{- 4} N$

$3 \times 10^{- 5} N$

answer is B.

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Detailed Solution

$\begin{array}{l} F = qVBsin θ \\ F = 1 .6 \times 10^{- 19} \times 20 \times 10^{7} \times 1 .5 \times \frac{1}{2} \\ = 2 .4 \times 10^{- 11} N \end{array}$

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