A charged particle carrying charge q = 1 μC moves in a uniform magnetic field with velocity $v_1= 10^6 m/s$ at an angle of 45° with the x-axis in the xy-plane and experiences a force $F_1 = 0.4\sqrt{2} N$ along the negative z-axis. When the same particle moves with velocity $v_2 =10^6 m/s$ along the z-axis, it experiences a force $F_2$ in the y-direction. Find the magnitude of the magnetic field

0.8 T

The correct answer is Option (1) → 0.8 T

Given:

Charge: $q = 1\ \mu\text{C} = 1 \times 10^{-6}\ \text{C}$

Velocity: $v_1 = v_2 = 10^6\ \text{m/s}$

Force: $F_1 = 0.4 \sqrt{2}\ \text{N}$ along -z axis

Angle with x-axis: $45^\circ$ in xy-plane

Magnetic force formula: $\vec{F} = q \vec{v} \times \vec{B}$

Let magnetic field $\vec{B} = B \hat{y}$ (assume direction along y). For $v_1$ in xy-plane at 45° with x-axis:

Velocity vector: $\vec{v}_1 = v_1(\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = v_1(\frac{\sqrt{2}}{2} \hat{i} + \frac{\sqrt{2}}{2} \hat{j})$

Force: $\vec{F}_1 = q \vec{v}_1 \times \vec{B} = q v_1 \frac{\sqrt{2}}{2} \hat{i} \times B \hat{j} = q v_1 \frac{\sqrt{2}}{2} B \hat{k}$

Given $\vec{F}_1$ along -z, magnitude: $F_1 = q v_1 \frac{\sqrt{2}}{2} B$

Substitute values:

$0.4 \sqrt{2} = 1 \times 10^{-6} \cdot 10^6 \cdot \frac{\sqrt{2}}{2} \cdot B$

$0.4 \sqrt{2} = 1 \cdot \frac{\sqrt{2}}{2} \cdot B$

$0.4 \sqrt{2} = \frac{\sqrt{2}}{2} B \Rightarrow B = 0.4 \cdot 2 = 0.8\ \text{T}$

∴ Magnitude of magnetic field = 0.8 T