An infinitely long conductor POR is bent to form a right angle as shown in figure. A current i flows through POR. The magnetic field due to this current carrying conductor at the point M is B1. 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 remaining unchanged. The magnetic field at M is now B2. The ratio B1/B2 is given by

B→1=μ0I4πR⊙, B→2=μ0I4πR+μ0I/24πR⊙ ⇒ B2=μ0I4πR32⊙ ⇒B1B2=23

An infinitely long conductor POR is bent to form a right angle as shown in figure. A current i flows through POR. The magnetic field due to this current carrying conductor at the point M is B1. 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 remaining unchanged. The magnetic field at M is now B2. The ratio B1/B2 is given by

B→1=μ0I4πR⊙, B→2=μ0I4πR+μ0I/24πR⊙ ⇒ B2=μ0I4πR32⊙ ⇒B1B2=23

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An infinitely long conductor POR is bent to form a right angle as shown in figure. A current i flows through POR. The magnetic field due to this current carrying conductor at the point M is B₁. 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 remaining unchanged. The magnetic field at M is now B₂. The ratio B₁/B₂ is given by

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Detailed Solution

${\vec{B}}_{1} = \frac{μ_{0} I}{4 π R} ⊙, {\vec{B}}_{2} = \frac{μ_{0} I}{4 π R} + \frac{μ_{0} I / 2}{4 π R} ⊙ \Rightarrow B_{2} = \frac{μ_{0} I}{4 π R} (\frac{3}{2}) ⊙ \Rightarrow |\frac{B_{1}}{B_{2}}| = \frac{2}{3}$

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