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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of \(2 \times 10^5 \mathrm{~ms}^{-1}\). When the electric field is switched off, the proton moves along a circular path of radius 2 cm . The magnitude of electric field is \(x \times 10^4 \mathrm{~N} / \mathrm{C}\). The value of \(x\) is _______ Take the mass of the proton \(=1.6 \times 10^{-27} \mathrm{~kg}\).

  1. A 4
  2. B 6
  3. C 2
  4. D 8
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Answer & Solution

Correct Answer

(C) 2

Step-by-step Solution

Detailed explanation

\begin{aligned} & B v q=E q \\ & E=B v \\ & r=\frac{v m}{B q} \\ & B=\frac{m v}{r q} \\ & E=\left(\frac{m v}{r q}\right) v=\frac{m v^2}{r q} \\ & =\frac{1.6 \times 10^{-27} \times 4 \times 10^{10}}{2 \times 10^{-2} \times 1.6 \times 10^{-19}} \\ & =2 \times 10^4 \mathrm{~N} /…

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