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Question: Answered & Verified by Expert
A $2 \mathrm{MeV}$ proton is moving perpendicular to a uniform magnetic field of 2.5 tes/a. The force on the proton is
PhysicsMagnetic Effects of CurrentJEE Main
Options:
  • A $2.5 \times 10^{-10} \mathrm{~N}$
  • B $7.6 \times 10^{-11} \mathrm{~N}$
  • C $2.5 \times 10^{-11} \mathrm{~N}$
  • D $7.6 \times 10^{-12} \mathrm{~N}$
Solution:
1382 Upvotes Verified Answer
The correct answer is: $7.6 \times 10^{-12} \mathrm{~N}$
\(\begin{aligned} & E_k=\frac{1}{2} m v^2\left\{v=2 \times 10^7 \mathrm{~m} / \mathrm{s}\right) \\ & F=q v B \sin 90=8 \times 10^{-12} N\end{aligned}\)

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