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JEE Main 2023 (25 Jan, Shift 1) · Magnetic Field of Straight Wires

Magnetic Field of Straight Wires question from JEE Main 2023 (25 Jan, Shift 1)

Q45JEE Main 2023 · 25 JAN · SHIFT 1Medium

Magnetic Field of Straight Wires

Match List I with List II

Choose the correct answer from the options given below:

List I (Current configuration)List II (Magnitude of Magnetic Field at point O)
A.I.B0=μ0I4πr[π+2]B_0 = \frac{\mu_0 I}{4\pi r}[\pi + 2]
B.II.B0=μ04IrB_0 = \frac{\mu_0}{4}\frac{I}{r}
C.III.B0=μ0I2πr[π−1]B_0 = \frac{\mu_0 I}{2\pi r}[\pi - 1]
D.IV.B0=μ0I4πr[π+1]B_0 = \frac{\mu_0 I}{4\pi r}[\pi + 1]
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Answer: (c)

  1. Use: semi-infinite wire ending at the foot gives μ0I4πr\frac{\mu_0 I}{4\pi r}, an infinite wire gives μ0I2πr\frac{\mu_0 I}{2\pi r}, a semicircle gives μ0I4r\frac{\mu_0 I}{4r}, a full loop gives μ0I2r\frac{\mu_0 I}{2r}; a wire whose line passes through O gives zero.
  2. Full loop with infinite wire opposing: μ0I2r−μ0I2πr=μ0I2πr[π−1]\frac{\mu_0 I}{2r} - \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{2\pi r}[\pi - 1] (III).
  3. Semicircle with two semi-infinite wires adding (hairpin): μ0I4πr[π+2]\frac{\mu_0 I}{4\pi r}[\pi + 2] (I).
  4. Semicircle with one semi-infinite wire adding: μ0I4πr[π+1]\frac{\mu_0 I}{4\pi r}[\pi + 1] (IV); semicircle with both wires through O: μ0I4r\frac{\mu_0 I}{4r} (II).
  5. So the answer is (c) A-III, B-I, C-IV, D-II.
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