AP EAMCET · PHYSICS · Magnetic Effects of Current
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
- A \(2.4 \times 10^{-12} \mathrm{~N}\)
- B \(24 \times 10^{-12} \mathrm{~N}\)
- C \(0.24 \times 10^{-12} \mathrm{~N}\)
- D \(0.024 \times 10^{-12} \mathrm{~N}\)
Answer & Solution
Correct Answer
(A) \(2.4 \times 10^{-12} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
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,…
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