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A long straight wire, carrying current I, is bent at its midpoint to form an angle of 45^o. The magnetic field at point P which is at distance R from the point of bending must be

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(2−1)μ0I4πR

(2+1)μ0I4πR

(2+1)μ0I42πR

(2−1)μ0I22πR

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detailed solution

Correct option is A

For segment BC, perpendicular distance isr=Rcos⁡45∘B and C ends are on the same side of the perpendicular PN.Therefore, α becomes negative and β is positive.Hence, α=−45∘ and β=90∘UsingB=μ0I4πr(Sin⁡α+Sin⁡β)⇒B=(2−1)μ0I4πR(into the plane of paper)

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An infinitely long conductor PQR is bent to form a right angle as shown. A current f flows through PQR. The magnetic field due to this current at the point M is H₁. Now another infinitely long straight conductor QS is connected at Q so that the current is I/2 in QR as well as in QS. The current in PQ remains unchanged. The magnetic field at M is now H₂. The ratio H₁/H₂ is given by