AP EAMCET · PHYSICS · Magnetic Effects of Current
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
- A \(3 \times 10^{-12} \mathrm{~N}\)
- B \(5 \times 10^{-12} \mathrm{~N}\)
- C \(6 \times 10^{-12} \mathrm{~N}\)
- D \(9 \times 10^{-12} \mathrm{~N}\)
Answer & Solution
Correct Answer
(B) \(5 \times 10^{-12} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
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…
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