Question 5
Solution
We analyze the work done on the particle in both cases. According to the Work-Energy Theorem, the change in kinetic energy is equal to the work done by all forces.
Case 1: Electric Field Only
The electric force \(F_E = qE\) acts in the upward direction (\(+\hat{j}\)). The particle deflects upwards by a distance \(y_1\). Work done is:
$$ W_1 = qE y_1 $$ $$ \frac{1}{2}m(v_1^2 – u^2) = W_1 $$Case 2: Electric + Magnetic Field
The magnetic field is into the page (\(-\hat{k}\) or similar depending on convention, but let’s use the cross product). Velocity is initially \(+\hat{i}\). Magnetic force \(\vec{F}_m = q(\vec{v} \times \vec{B})\).
Assuming standard orientation (B out of page or into page to oppose E):
- \(\vec{F}_E\) is Up.
- \(\vec{v}\) is Right. \(\vec{B}\) is out of page (dots in diagram). \(\vec{v} \times \vec{B}\) points Down.
The magnetic force opposes the electric force. The net upward force is reduced (or becomes downward). Consequently, the vertical deflection \(y_2\) is strictly less than \(y_1\).
Work Done:
Magnetic force does zero work. Only Electric force does work. Work depends on vertical displacement \(y_2\).
$$ W_2 = qE y_2 $$Since the upward force is reduced, the particle deflects less (\(y_2 < y_1\)). Therefore, \(W_2 < W_1\).
Conclusion:
Since less work is done in the second case, the final kinetic energy is lower.
$$ v_2 < v_1 $$This holds true regardless of the relationship between \(u\) and \(E/B\) (provided the particle enters the field).
Correct Answers: (a), (b), and (d)