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A proton moving with a velocity \(2.5 \times 10^7 \mathrm{~m} / \mathrm{s}\), enters a magnetic field of intensity \(2.5 \mathrm{~T}\) making an angle \(30^{\circ}\) with the magnetic field. The force on the proton is
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Verified Answer
The correct answer is:
\(5 \times 10^{-12} \mathrm{~N}\)
Velocity of proton, \(v=25 \times 10^7 \mathrm{~m} / \mathrm{s}\)
\(\text { Magnetic field, } \begin{aligned}
B & =2.5 \mathrm{~T} \\
\theta & =30^{\circ}
\end{aligned}\)
Magnetic force on proton in magnetic field is given as
\(\begin{aligned}
F & =B q v \sin \theta \\
& =2.5 \times 1.6 \times 10^{-19} \times 2.5 \times 10^7 \sin 30^{\circ} \\
& =6.25 \times 1.6 \times 10^{-12} \times \frac{1}{2}=5 \times 10^{-12} \mathrm{~N}
\end{aligned}\)
\(\text { Magnetic field, } \begin{aligned}
B & =2.5 \mathrm{~T} \\
\theta & =30^{\circ}
\end{aligned}\)
Magnetic force on proton in magnetic field is given as
\(\begin{aligned}
F & =B q v \sin \theta \\
& =2.5 \times 1.6 \times 10^{-19} \times 2.5 \times 10^7 \sin 30^{\circ} \\
& =6.25 \times 1.6 \times 10^{-12} \times \frac{1}{2}=5 \times 10^{-12} \mathrm{~N}
\end{aligned}\)
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