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 \(\vec{B} \) of magnitude 9 × 10⁻⁴ T perpendicular to its direction of motion. We wish to apply an uniform electric \(\vec{E} \) together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 × 10⁸ ms⁻¹)

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NEET 2025 Official Paper (Held On: 04 May, 2025)

\(\vec{E} \) is perpendicular to \(\vec{B} \) and
its magnitude is 27 × 10⁴ V m⁻¹
\(\vec{E} \) is perpendicular to \(\vec{B} \) and
its magnitude is 27 × 10² V m⁻¹
\(\vec{E} \) is parallel to \(\vec{B} \) and
its magnitude is 27 × 10² V m⁻¹
\(\vec{E} \) is parallel to \(\vec{B} \) and
its magnitude is 27 × 10⁴ V m⁻¹

Answer 1

Calculation:

The magnetic force on a moving charge is given by:

F_magnetic = q(v × B)

The electric force is given by:

F_electric = qE

For no deflection, the forces must balance:

F_electric = F_magnetic

qE = qv × B

Therefore, E = v× B which means E should be perpendicular to both v and B.

Now, the velocity (v) of the electron is:

v = c / 100 = (3 × 10⁸) / 100 = 3 × 10⁶ m/s

Substitute the values for B and v:

E = (3 × 10⁶) × (9 × 10^-4) = 27 × 10² V/m

Conclusion: The electric field (E) is perpendicular to the magnetic field (B) and its magnitude is 27 × 10² V/m.