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GUJCET · Physics · Moving Charges and Magnetism
A proton is moving perpendicular to a uniform magnetic field of 2.5 T with 2 Mev kinetic energy.
The force on proton is ________ N.
(Mass of proton \(=1.6 \times 10^{-27} kg\). charge of proton \(=1.6 \times 10^{-19} C\) )
- A \(8 \times 10^{-12}\)
- B \(8 \times 10^{-11}\)
- C \(3 \times 10^{-11}\)
- D \(3 \times 10^{-10}\)
Answer & Solution
Correct Answer
(A) \(8 \times 10^{-12}\)
Step-by-step Solution
Detailed explanation
C
Force on moving charge in magnetic field.
\(\begin{array}{l} F = B q \cup \sin \theta \\ F= B q \cup \quad\left(\because \theta=90^{\circ}\right)\end{array}\)
\(\begin{aligned} \therefore \quad F = B e \sqrt{\frac{2 K}{m}} \\ \left(\because q=e \text { and } v=\sqrt{\frac{2 K}{m}}\right)\end{aligned}\)
\(\begin{aligned} \therefore \quad F = 2.5 \times 1.6 \times 10^{-19} \times \sqrt{\frac{2 \times 2 \times 10^6 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27}}}\end{aligned}\)
\(\begin{array}{ll}\therefore F=2.5 \times 1.6 \times 10^{-19} \times 2 \times 10^7 \\ \therefore F=8 \times 10^{-12} N\end{array}\)
Force on moving charge in magnetic field.
\(\begin{array}{l} F = B q \cup \sin \theta \\ F= B q \cup \quad\left(\because \theta=90^{\circ}\right)\end{array}\)
\(\begin{aligned} \therefore \quad F = B e \sqrt{\frac{2 K}{m}} \\ \left(\because q=e \text { and } v=\sqrt{\frac{2 K}{m}}\right)\end{aligned}\)
\(\begin{aligned} \therefore \quad F = 2.5 \times 1.6 \times 10^{-19} \times \sqrt{\frac{2 \times 2 \times 10^6 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27}}}\end{aligned}\)
\(\begin{array}{ll}\therefore F=2.5 \times 1.6 \times 10^{-19} \times 2 \times 10^7 \\ \therefore F=8 \times 10^{-12} N\end{array}\)
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