An electron moving along the positive y-direction enters a region of uniform magnetic field \( \vec{B} \) directed along the positive z-direction. An electric field of suitable magnitude is also set up in the region so that the electron goes straight on without any deflection. The electric field must be directed along the
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → negative x-direction
The electron goes straight on only if the total Lorentz force \( \vec{F}=q(\vec{E}+\vec{v}\times\vec{B}) \) on it is zero. The electric force must then be equal and opposite to the magnetic force. For an electron (\( q=-e \)), both forces point opposite to what they would be for a positive charge.
With \( \vec{v}=v\,\hat{\jmath} \) and \( \vec{B}=B\,\hat{k} \), the magnetic force on the electron is:
\( \begin{aligned} \vec{F}_B &= -e\,(\vec{v}\times\vec{B}) \\ &= -e\,(v\,\hat{\jmath}\times B\,\hat{k}) \\ &= -evB\,\hat{\imath} \end{aligned} \)
The electric force \( -e\vec{E} \) has to be \( +evB\,\hat{\imath} \) to balance it, so
\( \vec{E}=-vB\,\hat{\imath} \)
Option (1) negative x-direction — Correct. A field along \( -\hat{\imath} \) pushes the negatively charged electron along \( +\hat{\imath} \), against the magnetic force along \( -\hat{\imath} \). With \( E=vB \) the two forces balance.
Option (2) positive x-direction — Incorrect. This is the direction needed for a positive charge. On an electron, the electric force acts opposite to \( \vec{E} \). So this field adds a second force along \( -\hat{\imath} \), and the electron bends even more.
Option (3) positive z-direction — Incorrect. A field along the z-axis gives a force along the z-axis. It cannot balance a magnetic force that acts along the x-axis.
Option (4) negative y-direction — Incorrect. This field lies along the line of motion. It pushes the electron along \( +\hat{\jmath} \) and only changes its speed, leaving the sideways magnetic force unbalanced.
NCERT (2026-27):
Magnetic Field, Lorentz Force – “… The force on an electric charge q due to both of them can be written as”
\( \vec{F}=q\,[\,\vec{E}(\vec{r})+\vec{v}\times\vec{B}(\vec{r})\,] \)\( \equiv\vec{F}_{\mathrm{electric}}+\vec{F}_{\mathrm{magnetic}} \qquad (4.3) \)
“… (i) It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge.”