Question

A current \(I_0\) flows through a metallic circular loop of radius \(r\) as shown. The resistance of arc \(ABC\) is half that of arc \(ADC\). Find the magnetic field at the centre \(O\).

• A. \(\frac{\mu_0 I_0}{6r}\)
• B. \(\frac{\mu_0 I_0}{2r}\)
• C. \(\frac{\mu_0 I_0}{12r}\)
• D. \(\frac{\mu_0 I_0}{4r}\)

Hint:

For parallel branches, current divides inversely proportional to resistance. Magnetic field due to a semicircle: \[ B=\frac{\mu_0 I}{4r} \] Always check whether the fields add or subtract.

Answer 1

The Correct Option is C

Concept: Current divides inversely proportional to resistance. Magnetic field at the centre due to a semicircular arc is \[ B=\frac{\mu_0 I}{4r} \]

Step 1: Find current division
Let resistance of arc \(ADC=R\). Then \[ R_{ABC}=\frac{R}{2} \] Current through \(ABC\), \[ I_{ABC} = I_0 \frac{R}{R+\frac{R}{2}} = \frac{2I_0}{3} \] Current through \(ADC\), \[ I_{ADC} = I_0 \frac{\frac{R}{2}}{R+\frac{R}{2}} = \frac{I_0}{3} \]

Step 2: Find magnetic fields due to both arcs
\[ B_1 = \frac{\mu_0}{4r} \left( \frac{2I_0}{3} \right) \] \[ B_1 = \frac{\mu_0 I_0}{6r} \] Similarly, \[ B_2 = \frac{\mu_0}{4r} \left( \frac{I_0}{3} \right) \] \[ B_2 = \frac{\mu_0 I_0}{12r} \]

Step 3: Determine resultant field
The currents flow through opposite semicircular paths. Hence fields are opposite. \[ B=B_1-B_2 \] \[ B= \frac{\mu_0 I_0}{6r} - \frac{\mu_0 I_0}{12r} \] \[ B= \frac{\mu_0 I_0}{12r} \] \[ \boxed{B=\frac{\mu_0 I_0}{12r}} \] Therefore, \[ \boxed{\text{Option (C)}} \]