The magnetic field at centre, P will be (a) (μ₀)/(4π) (b) (μ₀)/(π) (c) (μ₀)/(2π) (d) 4μ₀π

Answer: (c). P is midway, r = 2.5 m from each wire. Same-direction currents give opposite fields at P. B₁ = (μ₀ × 5)/(2π × 2.5) = (μ₀)/(π), B₂ = (μ₀ × 2.5)/(2π × 2.5) = (μ₀)/(2π) B = B₁ - B₂ = (μ₀)/(2π) So the answer is (c) (μ₀)/(2π).

The magnetic field at centre, P will be | NEET 2000 (AIPMT)

Q70NEET 2000 · AIPMTEasy

Magnetic Field of Straight Wires

The magnetic field at centre, P will be

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Answer: (c)

P is midway, $r = 2.5\text{ m}$ from each wire. Same-direction currents give opposite fields at P.
$B_1 = \frac{\mu_0 \times 5}{2\pi \times 2.5} = \frac{\mu_0}{\pi}$ , $B_2 = \frac{\mu_0 \times 2.5}{2\pi \times 2.5} = \frac{\mu_0}{2\pi}$
$B = B_1 - B_2 = \frac{\mu_0}{2\pi}$
So the answer is (c) $\frac{\mu_0}{2\pi}$ .

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