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Moving Charges and Magnetism question

2025 · Q172
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Moving Charges and Magnetism question

2025 · Q172

NEETPhysicsMoving Charges and MagnetismMCQ+4 / −1

An electron (mass 9×10−31 kg9 \times 10^{-31} \mathrm{~kg}9×10−31 kg and charge 1.6×10−19C1.6 \times 10^{-19} \mathrm{C}1.6×10−19C ) moving with speed c/100(c=c / 100(c=c/100(c= speed of light) is injected into a magnetic field B⃗\vec{B}B of magnitude 9×10−4 T9 \times 10^{-4} \mathrm{~T}9×10−4 T perpendicular to its direction of motion. We wish to apply an uniform electric field E⃗\vec{E}E together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c=3c=3c=3 ×103 ms−1\times 10^3 \mathrm{~ms}^{-1}×103 ms−1)

  1. A
    E⃗\vec{E}E is parallel to B⃗\vec{B}B and its magnitude is 27×102 V m−127 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1}27×102 V m−1
  2. B
    E⃗\vec{E}E is parallel to B⃗\vec{B}B and its magnitude is 27×104 V m−127 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}27×104 V m−1
  3. C
    E⃗\vec{E}E is perpendicular to B⃗\vec{B}B and its magnitude is 27×104 V m−127 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}27×104 V m−1
  4. D
    E⃗\vec{E}E is perpendicular to B⃗\vec{B}B and its magnitude is 27×102 V m−127 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1}27×102 V m−1
View written solutionFree

Correct answer: D

For no deflection of electron, $\vec{F}_B=\vec{F}_E$

NEET 2025 Physics - Moving Charges and Magnetism Question 1 English Explanation

$$\begin{aligned} \& -e(\vec{v} \times \vec{B})=-e \vec{E} \\ \& \Rightarrow \vec{E}=\vec{v} \times \vec{B} \Rightarrow \vec{E} \perp \vec{B} \\ \& E=v B=\frac{c}{100} \times 9 \times 10^{-4} \\ \& =\frac{3 \times 10^8}{100} \times 9 \times 10^{-4} \\ \& =27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1} \end{aligned}$$

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