CUET · PHYSICS · PYQ PAPER 2025
A charged particle carrying charge \(q = 1 \mu C\) moves in a uniform magnetic field with velocity \(v_1 = 10^6 \text{ m/s}\) at an angle of \(45^\circ\) with the x-axis in the xy-plane and experiences a force \(F_1 = 0.4\sqrt{2} \text{ N}\) along the negative z-axis. When the same particle moves with velocity \(v_2 = 10^6 \text{ m/s}\) along the z-axis, it experiences a force \(F_2\) in the y-direction. Find the magnitude of the magnetic field.
- A \(0.8 T\)
- B \(8 T\)
- C \(0.08 T\)
- D \(8 mT\)
Answer & Solution
Correct Answer
(A) \(0.8 T\)
Step-by-step Solution
Detailed explanation
From \(\vec{F_2} = q (\vec{v_2} \times \vec{B})\): Since \(\vec{v_2}\) is along z-axis and \(\vec{F_2}\) is along y-axis, \(\vec{B}\) must be along x-axis (\(\hat{k} \times \hat{i} = \hat{j}\)). For the first case, \(\vec{v_1}\) is at \(45^\circ\) to the x-axis in the xy-plane…
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