NEET 2025 — Physics solution

NEET2025Physics

An electron (mass $9 \times 1 0^{- 31}$ kg and charge $1.6 \times 1 0^{- 19}$ C) moving with speed $c /100$ ( $c$ = speed of light) is injected into a magnetic field $B$ of magnitude $9 \times 1 0^{- 4}$ T perpendicular to its direction of motion. We wish to apply a uniform electric field $E$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 1 0^{8}$ m s $^{- 1}$ ):

A
$E$ is perpendicular to $B$ and its magnitude is $27 \times 1 0^{4}$ V m $^{- 1}$
B
$E$ is perpendicular to $B$ and its magnitude is $27 \times 1 0^{2}$ V m $^{- 1}$
C
$E$ is parallel to $B$ and its magnitude is $27 \times 1 0^{2}$ V m $^{- 1}$
D
$E$ is parallel to $B$ and its magnitude is $27 \times 1 0^{4}$ V m $^{- 1}$

Solution

For the electron to travel undeflected, the electric force must exactly cancel the magnetic force: $F_{e} = - F_{m}$ . This requires $E = - v \times B$ , so $E$ must be perpendicular to both $v$ and $B$ (hence perpendicular to $B$ ).

Magnitude: $∣ E ∣ = v B sin 9 0^{\circ} = \frac{c}{100} \times B = \frac{3 \times 1 0 ^{8}}{100} \times 9 \times 1 0^{- 4} = 3 \times 1 0^{6} \times 9 \times 1 0^{- 4} = 27 \times 1 0^{2}$ V/m.

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