Questions

In the given figure, net magnetic field at O will be

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2μ0iπa4−π2

μ0i3πa4+π2

2μ0i3πa24+π2

2μ0i3πa4−π2

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detailed solution

Correct option is B

Magnetic field at O due toPart (1):B1=0Part(2):B2=μ04π⋅πi(a/2)⊗ ( along −Z-axis) Part(3):B3=μ04π⋅i(a/2)(↓) ( along −Y-axis) Part (4): B4=μ04π⋅πi(3a/2)⊙ (along +Z-axis) Part (5):B5=μ04π⋅i(3a/2)(↓) ( along −Y-axis) B2−B4=μ04π⋅πia2−23=μ0i3a⊗B3+B5=μ04π⋅ia2+23=8μ0i12πa(↓) ( along −Y-axis )Hence, net magnetic fieldBnet=B2−B42+B3+B52=μ0i3πaπ2+4

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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