JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2025
- Hard
- Numerical
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 \(=1.6 \times 10^{-27} \mathrm{~kg}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & B v q=E q \\ & E=B v \\ & r=\frac{v m}{B q} \\ & B=\frac{m v}{r q} \\ & E=\left(\frac{m v}{r q}\right) v=\frac{m v^2}{r q} \\ & =\frac{1.6 \times 10^{-27} \times 4 \times 10^{10}}{2 \times 10^{-2} \times 1.6 \times 10^{-19}} \\ & =2 \times 10^4 \mathrm{~N} / \mathrm{C} \\ & x=2\end{aligned}\)
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