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JEE Main 2019 (12 Apr, Shift 2) · Magnetic Field of Straight Wires

Magnetic Field of Straight Wires question from JEE Main 2019 (12 Apr, Shift 2)

Q59JEE Main 2019 · 12 APR · SHIFT 2Medium

Magnetic Field of Straight Wires

Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5A. (See figure) (μ\mu0 = 4π\pi × 10–7 N-A–2)

Figure

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Answer: (a)

  1. Perpendicular distance of P from AB: d=52−32=4d = \sqrt{5^2 - 3^2} = 4 cm; sin⁡α1=sin⁡α2=35\sin\alpha_1 = \sin\alpha_2 = \frac{3}{5}.
  2. B=μ0I4πd(sin⁡α1+sin⁡α2)=10−7×50.04×65B = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2) = \frac{10^{-7} \times 5}{0.04} \times \frac{6}{5}
  3. B=1.5×10−5B = 1.5 \times 10^{-5} T
  4. So the answer is (a).
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