A charged particle carrying charge \( q = 1\,\mu\mathrm{C} \) moves in a uniform magnetic field with velocity \( v_1 = 10^{6}\,\mathrm{m/s} \) at an angle of \( 45^\circ \) with the x-axis in the xy-plane and experiences a force \( F_1 = 0.4\sqrt{2}\,\mathrm{N} \) along the negative z-axis. When the same particle moves with velocity \( v_2 = 10^{6}\,\mathrm{m/s} \) along the z-axis, it experiences a force \( F_2 \) in the y-direction. Find the magnitude of the magnetic field
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.8 T
The magnetic force \( \vec{F} = q\,\vec{v}\times\vec{B} \) is perpendicular to both \( \vec{v} \) and \( \vec{B} \). This fixes the direction of the field first.
\( F_2 \) is along \( y \), so \( \vec{B} \) has no \( y \)-component.
\( F_1 \) is along \( -z \), so \( \vec{B} \) has no \( z \)-component.
Hence the field is along the x-axis, and the velocity \( v_1 \) makes \( 45^\circ \) with it.
\( \displaystyle F_1 = q\,v_1\,B\sin 45^\circ \;\Rightarrow\; B = \frac{F_1}{q\,v_1\sin 45^\circ} \)
\( \displaystyle q\,v_1 = 10^{-6} \times 10^{6} = 1\,\mathrm{N\,T^{-1}} \)
\( \displaystyle B = \frac{0.4\sqrt{2}}{1 \times (1/\sqrt{2})} = 0.4\sqrt{2} \times \sqrt{2} = 0.8\,\mathrm{T} \)