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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

  1. A \(3 \times 10^{-12} \mathrm{~N}\)
  2. B \(5 \times 10^{-12} \mathrm{~N}\)
  3. C \(6 \times 10^{-12} \mathrm{~N}\)
  4. D \(9 \times 10^{-12} \mathrm{~N}\)
Verified Solution

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…