Physics Q3 – NEET-UG 2025

Home Physics Moving ChargesQuestion #350

An electron (mass

9 \times 10^{- 31}

kg and charge

1.6 \times 10^{- 19}

C) moving with speed

c / 100

(

c = speed of light

) is injected into a magnetic field

\vec{B}

of magnitude

9 \times 10^{- 4}

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

c = 3 \times 10^{8}

^{- 1}

)

Clue 1 of 3: Velocity Selector Principle

For a charged particle to pass undeflected through a region with both electric and magnetic fields, the net force acting on it must be zero.

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● Video Walkthrough

Step 1: Apply the Condition for No Deflection✦ Active

For an electron to pass undeflected through combined electric and magnetic fields, the net force on it must be zero. This means the electric force must balance the magnetic force. The Lorentz force equation is $\vec{F} = q \vec{E} + q (\vec{v} \times \vec{B})$ . For no deflection, $\vec{F} = 0$ , so $q \vec{E} = - q (\vec{v} \times \vec{B})$ . Since $q \neq 0$ , we have $\vec{E} = - (\vec{v} \times \vec{B})$ . This implies that $\vec{E}$ must be perpendicular to both $\vec{v}$ and $\vec{B}$ .

💡 Teacher's Secret Hint

Remember that for a velocity selector, the electric and magnetic forces must be equal in magnitude and opposite in direction.

Step 2: Calculate the Magnitude of the Electric Field○ Expand

Given that the magnetic field $\vec{B}$ is perpendicular to the velocity $\vec{v}$ , the magnitude of the magnetic force is $F_{B} = | q | v B$ . For the forces to balance, the magnitude of the electric force $F_{E} = | q | E$ must be equal to $F_{B}$ . Thus, $| q | E = | q | v B$ , which simplifies to $E = v B$ .

First, calculate the speed of the electron $v$ :

v = \frac{c}{100} = \frac{3 \times 10^{8} m/s}{100} = 3 \times 10^{6} m/s

Now, calculate the magnitude of the electric field $E$ :

E = v B = (3 \times 10^{6} m/s) \times (9 \times 10^{- 4} T)

E = 27 \times 10^{(6 - 4)} V/m = 27 \times 10^{2} V/m

💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation. The unit for electric field is V/m or N/C.

Step 3: Determine the Correct Option○ Expand

From Step 1, we established that $\vec{E}$ must be perpendicular to $\vec{B}$ . This eliminates options (3) and (4). From Step 2, we calculated the magnitude of the electric field to be $27 \times 10^{2}$ V m $^{- 1}$ . Comparing this with the remaining options, option (2) matches both the direction and magnitude.

💡 Teacher's Secret Hint

Always check both the magnitude and direction when dealing with vector quantities in options.

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Physics Question 3 – NEET-UG 2025

Clue 1 of 3: Velocity Selector Principle

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