In the given figure MN and OP are semi-infinite wires and segment NO is a semi-ring. The magnetic field at C is (in T)

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Solution

The correct option is C

$We have B_{M N} = 0 B_{N O} = \frac{μ_{0} i}{4 R} \otimes B_{O P} = \frac{μ_{0} i}{4 π R} \otimes B = B_{M N} + B_{N O} + B_{O P} = \frac{μ_{0} i}{4 π R} (π + 1) \otimes$

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In the given figure MN and OP are semi-infinite wires and segment NO is a semi-ring. The magnetic field at C is (in T)

The correct option is C We haveBMN=0BNO=μ0i4R⊗BOP=μ0i4πR⊗B=BMN+BNO+BOP=μ0i4πR(π+1)⊗

The correct option is C

$We have B_{M N} = 0 B_{N O} = \frac{μ_{0} i}{4 R} \otimes B_{O P} = \frac{μ_{0} i}{4 π R} \otimes B = B_{M N} + B_{N O} + B_{O P} = \frac{μ_{0} i}{4 π R} (π + 1) \otimes$