Question

Two circular coils X (smaller) and Y (bigger), each having a single turn, carry equal currents in the same direction and subtend the same angle at point 'O' along the axis of the coil. The distances of the centers of coil to point 'O' are \(d\) and \((\frac{d}{2})\) for coil Y and X respectively. The radii of coils Y and X are \((2r)\) and \((r)\) respectively. The magnetic induction due to the bigger coil at point 'O' is \(B_y\) and that due to smaller coil X at point 'O' is \(B_x\). (\(d≫r\)) The relation between \(B_x\) and \(B_y\) is

• A. \(B_y = B_x\)
• B. \(B_y = 2B_x\)
• C. \(B_x = 2B_y\)
• D. \(B_x = 4B_y\)

Hint:

Use the axial field formula and the condition d much larger than r.

Answer 1

The Correct Option is C

Step 1: Axial Field of a Circular Coil:
\[ B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}} \]
For \(x\gg R\) this becomes \(B\approx\dfrac{\mu_0IR^2}{2x^3}\). The condition \(d\gg r\) lets us use this form for both coils.

Step 2: Coil Y:
Radius \(2r\), distance \(d\): \(B_y=\dfrac{\mu_0I(2r)^2}{2d^3}=\dfrac{4\mu_0Ir^2}{2d^3}\).

Step 3: Coil X:
Radius \(r\), distance \(d/2\): \(B_x=\dfrac{\mu_0Ir^2}{2(d/2)^3}=\dfrac{8\mu_0Ir^2}{2d^3}\).

Step 4: Compare:
\[ \frac{B_x}{B_y}=\frac84=2\Rightarrow B_x=2B_y \]
This is option (C). The two coils subtend the same angle at O because \(\dfrac{2r}{d}=\dfrac r{d/2}\), but that does not make the fields equal, since the field also depends on the distance.

Final Answer:
\(B_x=2B_y\), option (C). \[ \boxed{\text{(C) } B_x=2B_y} \]