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JEE Main 2022 (27 Jun, Shift 2) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2022 (27 Jun, Shift 2)

Q62JEE Main 2022 · 27 JUN · SHIFT 2Difficult

Moving Charge in a Magnetic Field

Two long parallel conductors S1 and S2 are separated by a distance 10 cm and carrying currents of 4A and 2A respectively. The conductors are placed along x-axis in X-Y plane. There is a point P located between the conductors (as shown in figure).

A charged particle of 3π\pi coulomb is passing through the point P with velocityv⃗=(2i^+3j^)\vec{v} = (2\hat{i} + 3\hat{j}) m/s; where i^\hat{i} and j^\hat{j} represents unit vector along x & y axis respectively.

The force acting on the charged particle is 4π×10−5(−xi^+2j^)4\pi \times 10^{-5}(-x\hat{i} + 2\hat{j}) N. The value of x is:

Figure

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Answer: (c)

  1. PP is 4 cm below S1S_1 and 6 cm above S2S_2; both currents are along +x+x.
  2. From S1S_1: B1=μ0(4)2π(0.04)=2×10−5 TB_1 = \dfrac{\mu_0 (4)}{2\pi(0.04)} = 2 \times 10^{-5}\text{ T} along −k^-\hat{k}. From S2S_2: B2=μ0(2)2π(0.06)=23×10−5 TB_2 = \dfrac{\mu_0 (2)}{2\pi(0.06)} = \tfrac{2}{3} \times 10^{-5}\text{ T} along +k^+\hat{k}.
  3. Net B⃗=−43×10−5 k^ T\vec{B} = -\tfrac{4}{3} \times 10^{-5}\,\hat{k}\text{ T}.
  4. F⃗=qv⃗×B⃗=3π(2i^+3j^)×(−43×10−5k^)=4π×10−5(−3i^+2j^) N\vec{F} = q\vec{v} \times \vec{B} = 3\pi(2\hat{i} + 3\hat{j}) \times (-\tfrac{4}{3} \times 10^{-5}\hat{k}) = 4\pi \times 10^{-5}(-3\hat{i} + 2\hat{j})\text{ N}
  5. So x=3x = 3, answer (c).
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