Question

A single turn circular coil is connected to a cell as shown. Magnetic field at the centre \(O\) of the coil is

• A. \(\frac{2\pi I}{r}\)
• B. \(2\pi Ir\)
• C. zero
• D. \(\frac{I}{2\pi r}\)
• E. \(\frac{I}{\pi r}\)

Hint:

Whenever current splits symmetrically in opposite directions in a loop, magnetic fields at symmetric points cancel.

Answer 1

The Correct Option is C

Concept: Magnetic field at the centre of a circular loop depends on direction of current and symmetry. If equal currents flow in opposite directions, their magnetic effects cancel.

Step 1: Understand current flow in the circuit.
The given circuit shows a circular loop connected to a battery. The connection is such that current enters the loop and splits into two equal paths along the circular wire.

Step 2: Nature of current distribution.
Since the loop is symmetric and resistance along both paths is identical, the current divides equally: \[ I \rightarrow \frac{I}{2} \text{ in each half} \]

Step 3: Direction of currents in the loop.
The two halves of the circular loop carry current in opposite directions (one clockwise, the other anticlockwise).

Step 4: Magnetic field due to each half.
Magnetic field at the centre due to a current-carrying arc: \[ B = \frac{\mu_0 I \theta}{4\pi r} \] For semicircle (\(\theta = \pi\)): \[ B = \frac{\mu_0 I}{4r} \] Each half produces equal magnitude magnetic field.

Step 5: Direction of magnetic fields.
Using right-hand rule:
• One semicircle produces field into the page
• Other produces field out of the page

Step 6: Net magnetic field.
Since magnitudes are equal and directions opposite: \[ B_{\text{net}} = B - B = 0 \]

Step 7: Final conclusion. \[ \boxed{0} \]