JEE Advanced Physics Question (2026) — Solution

Step-by-step solution

For the time interval 0 t 0.2 s, the particle is under the influence of the electric field and gravity. The charge-to-mass ratio of the particle is q m = 10^ -6 10^ -6 = 1 C kg^ -1 . The acceleration of the particle is: a = q E m + g = 1( i ) - 10 j = i - 10 j ms^ -2 Given the initial velocity v _0 = i + 2 j ms^ -1 and initial vertical position y_0 = 0 (since it is projected from the XZ plane), we can find the velocity and position at t = 0.2 s. Velocity at t = 0.2 s: v (0.2) = v _0 + a t = ( i + 2 j ) + ( i - 10 j )(0.2) = 1.2 i ms^ -1 Vertical position at t = 0.2 s: y(0.2) = y_0 + v_ 0y t + 1 2 a_y t^2 = 0 + 2(0.2) - 1 2 (10)(0.2)^2 = 0.4 - 0.2 = 0.2 m = 20 cm For t > 0.2 s, the electric field is switched off and the magnetic field B = 6 j T is switched on. The magnetic force F _m = q( v B ) acts only in the XZ plane because B is along the y-axis. The vertical motion is only affected by gravity. Let t' = t - 0.2 be the time elapsed after the fields are switched. The initial vertical velocity for this phase is v_y(0.2) = 0. The vertical position as a function of t' is: y(t') = y(0.2) + v_y(0.2)t' - 1 2 gt'^2 = 0.2 - 5t'^2 Evaluating the options: At t = 0.3 s (t' = 0.1 s): y(0.1) = 0.2 - 5(0.1)^2 = 0.2 - 0.05 = 0.15 m = 15 cm. (Option A is correct) At t = 0.4 s (t' = 0.2 s): y(0.2) = 0.2 - 5(0.2)^2 = 0.2 - 0.2 = 0 m = 0 cm. (Option B is incorrect) At t = 0.35 s (t' = 0.15 s): y(0.15) = 0.2 - 5(0.15)^2 = 0.2 - 0.1125 = 0.0875 m 0. (Option D is incorrect) For the radius of the trajectory for t > 0.2 s, the particle moves in a helical path. The radius of the circular projection in the XZ plane is determined by the velocity perpendicular to the magnetic field, v_ = 1.2 ms^ -1 . R = mv_ qB = 10^ -6 1.2 10^ -6 6 = 0.2 m = 20 cm. (Option C is correct) Answer: The vertical distance of the particle from the XZ plane at t = 0.3 s is 15 cm.; The radius of the trajectory of the particle for t > 0.2 s is 20 cm.