An electron is moving at a velocity of 100 m/s along the X-axis in the presence of an electric field of 30 V/m directed along the Y-axis and a magnetic field of 300 G directed along the Z-axis. The magnitude and direction of force acting on the electron is (Given: the charge on the electron is "e")
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 27 e N along the negative Y-axis
Given:
Velocity of electron: $\vec{v} = 100\,\text{m/s} \,\hat{i}$
Electric field: $\vec{E} = 30\,\text{V/m} \,\hat{j}$
Magnetic field: $\vec{B} = 300\,\text{G} = 300 \times 10^{-4}\,\text{T} = 0.03\,\text{T} \,\hat{k}$
Charge of electron: $q = -e$
Force on a charged particle:
$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$
Compute $\vec{v} \times \vec{B}$:
$\vec{v} \times \vec{B} = (100\,\hat{i}) \times (0.03\,\hat{k}) = 100 \cdot 0.03 \cdot (\hat{i} \times \hat{k}) = 3 \cdot (-\hat{j}) = -3\,\hat{j}$
So, $\vec{E} + \vec{v} \times \vec{B} = 30\,\hat{j} + (-3\,\hat{j}) = 27\,\hat{j}$
Then, $\vec{F} = -e \cdot (27\,\hat{j}) = -27e\,\hat{j}$
Magnitude of force:
$|\vec{F}| = 27e$
Direction: Along the negative Y-axis (i.e., $-\hat{j}$)