Question

Two concentric coplanar circular loops of radii $r_1$ and $r_2$ respectively carry currents $i_1$ and $i_2$ in opposite directions (one clockwise and other anticlockwise). The magnetic induction at the centre of the loops is half that due to $i_1$ alone at the centre. If $r_2 = 2r_1$, the value of $\frac{i_2}{i_1}$ is

• A. $\frac{1}{4}$
• B. $1$
• C. $2$
• D. $\frac{1}{2}$

Hint:

Think of it using the proportionality of magnetic fields: $B \propto \frac{i}{r}$. For the net field to drop to half of the first field ($1 - \Delta = 0.5$), the second loop's field must contribute exactly half the strength of the first ($B_2 = 0.5 B_1$). Since the second loop has twice the radius ($r_2 = 2r_1$), it naturally dilutes its field by half. To maintain $B_2 = 0.5 B_1$, its current must be exactly equal to the first loop's current!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Two concentric current loops produce a net magnetic field that is half the field of the first loop alone. Find the current in the second loop given its radius is double the first.

Step 2: Key Formula or Approach:
Magnetic field at the center of a circular loop B ∝ i/r. For the net field to be B₁ - B₂ = 0.5B₁, the second field must be B₂ = 0.5B₁.

Step 3: Detailed Explanation:
The second loop has twice the radius (r₂ = 2r₁), which by itself would halve its field contribution for equal current. To achieve exactly B₂ = 0.5B₁, the radius dilution already provides the needed factor of one-half. Therefore, the current in the second loop must equal the current in the first loop—no additional current scaling is required.

Step 4: Final Answer:
The second loop's current equals the first loop's current.