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:
The question describes two concentric, coplanar circular loops with currents flowing in opposite directions. The net magnetic field at their common center is exactly half of the magnetic field produced by the first loop alone. Given the relationship between their radii, we need to find the ratio of their currents, $\frac{i_2}{i_1}$.

Step 2: Key Formula or Approach:
1. The magnetic induction ($B$) at the center of a circular loop of radius $r$ carrying current $i$ is: $$B = \frac{\mu_0 i}{2r}$$ 2. Since the currents flow in opposite directions, the individual magnetic field vectors oppose each other: $$B_{\text{net}} = |B_1 - B_2|$$ 3. We are given $B_{\text{net}} = \frac{1}{2}B_1$. Depending on which field is larger, this sets up two possibilities: $B_1 - B_2 = \frac{1}{2}B_1$ or $B_2 - B_1 = \frac{1}{2}B_1$. Let's analyze the standard case where $B_1 > B_2$.

Step 3: Detailed Explanation:
Let's express the magnetic fields of both loops at the center: $$B_1 = \frac{\mu_0 i_1}{2r_1}$$ $$B_2 = \frac{\mu_0 i_2}{2r_2} = \frac{\mu_0 i_2}{2(2r_1)} = \frac{\mu_0 i_2}{4r_1}$$ Using the condition $B_1 - B_2 = \frac{1}{2}B_1$: $$B_1 - \frac{1}{2}B_1 = B_2 \implies \frac{1}{2}B_1 = B_2$$ Now, substitute the expressions for $B_1$ and $B_2$ into this relationship: $$\frac{1}{2} \left(\frac{\mu_0 i_1}{2r_1}\right) = \frac{\mu_0 i_2}{4r_1}$$ $$\frac{\mu_0 i_1}{4r_1} = \frac{\mu_0 i_2}{4r_1}$$ Canceling out the common factor $\frac{\mu_0}{4r_1}$ from both sides leaves: $$i_1 = i_2 \implies \frac{i_2}{i_1} = 1$$ (Note: The alternative case $B_2 - B_1 = \frac{1}{2}B_1 \implies B_2 = \frac{3}{2}B_1$ yields $\frac{i_2}{4r_1} = \frac{3\mu_0 i_1}{4r_1} \implies \frac{i_2}{i_1} = 3$, which is not present in the options).

Step 4: Final Answer:
The ratio of the currents $\frac{i_2}{i_1}$ is $1$, which corresponds to option (B).