An electron (mass 9 × 10^ -31 ~kg and charge 1.6 × 10^ -19 C ) moving with speed c / 100(c= speed of light) is injected into a magnetic field B of magnitude 9 × 10^ -4 ~T perpendicular to its direction of motion. We

Correct answer: B. E is perpendicular to B and its magnitude is 27 × 10^2 Vm^ -1 For particle To Remain undeflected aligned & F_e=- F_ m & | E |=VB θ & E= c 100 × 9 × 10^ -4 90 & E=27 × 10^2 V / m and & E = B × V aligned

NEET 2025

Exam: NEET
Year: 2025
Subject: Physics
Chapter: Magnetic Effects of Current

NEET 2025 · Physics · Magnetic Effects of Current

An electron (mass 9 × 10^ -31 ~kg and charge 1.6 × 10^ -19 C )...

An electron (mass

9 \times 10^{-31} \mathrm{~kg}

and charge

1.6 \times 10^{-19} \mathrm{C}

) moving with speed

\mathrm{c} / 100(\mathrm{c}=

speed of light) is injected into a magnetic field

\vec{B}

of magnitude

9 \times 10^{-4} \mathrm{~T}

perpendicular to its direction of motion. We wish to apply an uniform electric field

\vec{E}

together with the magnetic field so that the electron does not deflect from its path. Then (speed of light

\mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}

)

Options

A $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}$
B $\overrightarrow{\mathrm{E}}$ is perpendicular to $\overrightarrow{\mathrm{B}}$ and its magnitude is $27 \times 10^2 \mathrm{Vm}^{-1}$
C $\overrightarrow{\mathrm{E}}$ is parallel to $\overrightarrow{\mathrm{B}}$ and its magnitude is $27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1}$
D $\overrightarrow{\mathrm{E}}$ is parallel to $\overrightarrow{\mathrm{B}}$ and its magnitude is $27 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}$

Correct answer

B. $\overrightarrow{\mathrm{E}}$ is perpendicular to $\overrightarrow{\mathrm{B}}$ and its magnitude is $27 \times 10^2 \mathrm{Vm}^{-1}$

Step-by-step solution

For particle To Remain undeflected

\begin{aligned} & \overrightarrow{\mathrm{F}_e=-\overrightarrow{\mathrm{F}_{\mathrm{m}} \\ & \Rightarrow |\overrightarrow{\mathrm{E}}|=\mathrm{VB} \sin \theta \\ & \mathrm{E}=\frac{\mathrm{c}}{100} \times 9 \times 10^{-4} \sin 90 \\ & \mathrm{E}=27 \times 10^2 \mathrm{ V} / \mathrm{m} \text{ and } \\ & \overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{V}} \end{aligned}

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