A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2 × 10^5 ~ms^ -1 . When the electric field is switched off, the proton moves along a circular path of radius 2 cm

Correct answer: 0 aligned & B v q=E q & E=B v & r=(v m)/(B q) & B=(m v)/(r q) & E=((m v)/(r q)) v=(m v^2)/(r q) & = 1.6 × 10^ -27 × 4 × 10^ 10 2 × 10^ -2 × 1.6 × 10^ -19 & =2 × 10^4 ~N / C & x=2 aligned

JEE Main 202522 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Evening Shift
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Main 2025 · 22 January · Evening Shift · Physics · Magnetic Effects of Current

A proton is moving undeflected in a region of crossed electric and...

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}

Correct answer

Step-by-step solution

\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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