Question

In a particular case \(X\), a certain length of insulated copper wire is bent to form double loops of equal radii. In another case \(Y\) the same copper wire is bent to form three loops of equal radii. If the same steady current is passed through the copper wire in both the cases, the ratio of magnetic field at the centre in the case \(X\) to that in case \(Y\) is:

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{9}\)
• C. \(\frac{9}{4}\)
• D. \(\frac{4}{9}\)

Hint:

For the same wire length and same current, if a coil has \(N\) turns, then its radius becomes inversely proportional to \(N\), and magnetic field becomes proportional to \(N^2\).

Answer 1

The Correct Option is D

Step 1: Use magnetic field at the centre of circular coil.
For a circular coil of \(N\) turns and radius \(r\), magnetic field at the centre is:
\[ B = \frac{\mu_0 N I}{2r} \]

Step 2: Relate radius with number of loops.
The same length of wire is used in both cases. If total length is \(L\), then:
\[ L = N(2\pi r) \]
So,
\[ r = \frac{L}{2\pi N} \]

Step 3: Substitute radius in magnetic field formula.
\[ B = \frac{\mu_0 N I}{2 \left(\frac{L}{2\pi N}\right)} \]

Step 4: Simplify the expression.
\[ B = \frac{\mu_0 N I \times 2\pi N}{2L} \]
\[ B = \frac{\mu_0 \pi I N^2}{L} \]
Thus, for the same wire length and same current:
\[ B \propto N^2 \]

Step 5: Apply for case \(X\).
In case \(X\), double loops are formed, so:
\[ N_X = 2 \]
\[ B_X \propto 2^2 = 4 \]

Step 6: Apply for case \(Y\).
In case \(Y\), three loops are formed, so:
\[ N_Y = 3 \]
\[ B_Y \propto 3^2 = 9 \]

Step 7: Find required ratio.
\[ \frac{B_X}{B_Y} = \frac{4}{9} \]
\[ \boxed{\frac{4}{9}} \]