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JEE Main 2026 · 2 Apr · Shift 2, Physics question 29

A particle having charge 10−910^{-9} C moving in xx-yy plane in fields of 0.4j^0.4\hat j N/C and 4×10−3k^4\times10^{-3}\hat k T experiences a force of (4i^+2j^)×10−10\left(4\hat i+2\hat j\right)\times10^{-10} N. The velocity of the particle at that instant is ________ m/s.

Choose one option
Answer and worked solution

Official answer: A

Lorentz force. F⃗=q(E⃗+v⃗×B⃗)\vec F=q(\vec E+\vec v\times\vec B). With v⃗=(vx,vy,0)\vec v=(v_x,v_y,0) and B⃗=Bk^\vec B=B\hat k: v⃗×B⃗=B(vy,−vx,0)\vec v\times\vec B=B(v_y,-v_x,0).

Components.

  • xx: 10−9(4×10−3)vy=4×10−1010^{-9}(4\times10^{-3})v_y=4\times10^{-10}, so vy=100v_y=100.
  • yy: 10−9(0.4)−4×10−12vx=2×10−1010^{-9}(0.4)-4\times10^{-12}v_x=2\times10^{-10}, so vx=50v_x=50.

Velocity. v⃗=50i^+100j^\vec v=50\hat i+100\hat j m/s.

(A)

Topics:Motion in magnetic and electric fieldsForce on a moving charge
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