The magnetic field at the centre of semi-circular wire carrying current i is

  1. \(\frac{\mu_0i}{2r}\)
  2. \(\frac{\mu_0i}{4r}\)
  3. \(\frac{\mu_0i}{r}\)
  4. \(\frac{\mu_0i}{2\pi r}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{\mu_0i}{4r}\)
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Detailed Solution

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CONCEPT:

  • The Current-Carrying Conductor produces a magnetic Field Around it. 
    • Biot Savarts Law: The direction and Magnitude of this magnetic field can be obtained from Biot Savarts Law. 

\(\overrightarrow{dB} = \frac{μ _{0}\overrightarrow{I}\times \overrightarrow{dl}}{4\pi r^{2}}\)

Where dB = Magnetic Field produced due to small wire of length dl, I = Current in wire, μ= Permittivity of free space, dl = Small Current Element

  • The magnetic field generated due to the current-carrying circular conductor at its center is given as:

\(\Rightarrow B=\dfrac{\mu_0 i}{2r}\)

  • This result has been obtained from Biot Savarts Law.

EXPLANATION:

  • The Magnetic Field Produced due to the Semi-Circular wire is half of that of Circular wire.
  • So, the magnetic field produced at the center of the wire 

\(\Rightarrow B = \frac{1}{2}\times \dfrac{\mu_0 i}{2r}=\dfrac{\mu_0 i}{4r}\)

  • Hence option 2 is the correct option.
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