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JEE Main 2026 (2 Apr, Shift 2) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2026 (2 Apr, Shift 2)

Q13JEE Main 2026 · 2 APR · SHIFT 2Medium

Moving Charge in a Magnetic Field

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

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Answer: (a)

  1. F⃗=q(E⃗+v⃗×B⃗)\vec{F} = q(\vec{E} + \vec{v} \times \vec{B}); electric part qE=10−9×0.4 j^=4×10−10 j^qE = 10^{-9} \times 0.4\,\hat{j} = 4 \times 10^{-10}\,\hat{j} N.
  2. Magnetic part =(4i^+2j^−4j^)×10−10=(4i^−2j^)×10−10= (4\hat{i} + 2\hat{j} - 4\hat{j}) \times 10^{-10} = (4\hat{i} - 2\hat{j}) \times 10^{-10} N.
  3. With v⃗=vxi^+vyj^\vec{v} = v_x\hat{i} + v_y\hat{j}: qv⃗×Bk^=qB(vyi^−vxj^)q\vec{v} \times B\hat{k} = qB(v_y\hat{i} - v_x\hat{j}), qB=4×10−12qB = 4 \times 10^{-12}.
  4. 4×10−12vy=4×10−10⇒vy=1004 \times 10^{-12}v_y = 4 \times 10^{-10} \Rightarrow v_y = 100; 4×10−12vx=2×10−10⇒vx=504 \times 10^{-12}v_x = 2 \times 10^{-10} \Rightarrow v_x = 50
  5. So the answer is (a) 50i^+100j^50\hat{i} + 100\hat{j} m/s.
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