AP EAMCET · PHYSICS · Magnetic Effects of Current
If a proton of kinetic energy 8.35 MeV enters a uniform magnetic field of 10 T at right angles to the direction of the field, then the force acting on the proton is (Mass of proton \(=1.67 \times 10^{-27} \mathrm{~kg}\) and charge of proton \(=1.6 \times 10^{-19} \mathrm{C}\) )
- A \(48 \times 10^{-12} \mathrm{~N}\)
- B \(16 \times 10^{-12} \mathrm{~N}\)
- C \(64 \times 10^{-12} \mathrm{~N}\)
- D \(32 \times 10^{-12} \mathrm{~N}\)
Answer & Solution
Correct Answer
(C) \(64 \times 10^{-12} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\( K = 8.35 \, \text{MeV} = 8.35 \times 1.6 \times 10^{-13} \, \text{J} = 1.336 \times 10^{-12} \, \text{J} \) \( v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 1.336 \times 10^{-12}}{1.67 \times 10^{-27}}} = 4 \times 10^7 \, \text{m/s} \)…
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