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MHT CET 2023 (9 May, Shift 2) · Magnetic Field of Straight Wires

Magnetic Field of Straight Wires question from MHT CET 2023 (9 May, Shift 2)

Q37MHT CET 2023 · 9 MAY · SHIFT 2Easy

Magnetic Field of Straight Wires

The magnetic field at a point PP situated at perpendicular distance 'RR' from a long straight wire carrying a current of 12 A12\text{ A} is 3×10−5 Wb/m23 \times 10^{-5}\text{ Wb/m}^2. The value of 'RR' in mm\text{mm} is [μ0=4π×10−7 Wb/Am]\left[\mu_0 = 4\pi \times 10^{-7}\text{ Wb/Am}\right]

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Answer: (d)

  1. B=μ0I2πR⇒R=μ0I2πBB = \dfrac{\mu_0 I}{2\pi R} \Rightarrow R = \dfrac{\mu_0 I}{2\pi B}
  2. R=2×10−7×123×10−5=8×10−2R = \dfrac{2\times10^{-7}\times12}{3\times10^{-5}} = 8\times10^{-2} m
  3. R=80R = 80 mm
  4. So the answer is (d) 80.
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