A proton enters a magnetic field of flux density \(1.5 \mathrm{~Wb} \mathrm{~m}^{-2}\) with a velocity of…
A proton enters a magnetic field of flux density \(1.5 \mathrm{~Wb} \mathrm{~m}^{-2}\) with a velocity of \(2 \times 10^7 \mathrm{~ms}^{-1}\) at an angle of \(30^{\circ}\) with the field. The force on the proton will be
\(2.4 \times 10^{-12} \mathrm{~N}\)
\(24 \times 10^{-12} \mathrm{~N}\)
\(0.24 \times 10^{-12} \mathrm{~N}\)
\(0.024 \times 10^{-12} \mathrm{~N}\)
Solution
Magnetic flux density, \(B=1.5 \mathrm{Wbm}^{-2}\)
Velocity of proton, \(v=2 \times 10^7 \mathrm{~ms}^{-1}\)
\(\theta=30^{\circ}\)
Charge on proton, \(q=1.6 \times 10^{-19} \mathrm{C}\)
\(\therefore\) Force on the proton,
\(\begin{aligned}
F & =B q v \sin \theta \\
& =1.5 \times 1.6 \times 10^{-19} \times 2 \times 10^7 \times \sin 30^{\circ} \\
& =4.8 \times 10^{-12} \times \frac{1}{2}=2.4 \times 10^{-12} \mathrm{~N}
\end{aligned}\)