Question

The magnetic moments associated with two closely wound circular coils $A$ and $B$ of radius $r _{ A }=10$ $cm$ and $r _B=20 cm$ respectively are equal if : (Where $N _A, I _A$ and $N _B, I _{ B }$ are number of turn and current of $A$ and $B$ respectively)

• A. $2 N _{ A } I _{ A }= N _{ B } I _{ B }$
• B. $N _{ A }=2 N _{ B }$
• C. $N _{ A } I _{ A }=4 N _{ B } I _{ B }$
• D. $4 N _{ A } I _{ A }= N _{ B } I _{ B }$

Answer 1

The Correct Option is C

The question revolves around the concept of magnetic moments in circular coils. The magnetic moment (\(M\)) for a circular coil is given by the formula:

\(M = N \cdot I \cdot A\)

where:

• \(N\) is the number of turns
• \(I\) is the current flowing through the coil
• \(A\) is the area of the coil

The area \(A\) of a circular coil is given by \(\pi r^2\), where \(r\) is the radius of the coil.

Let's analyze the coiled circles A and B:

• Radius of coil A: \(r_A = 10 \, \text{cm}\)
• Radius of coil B: \(r_B = 20 \, \text{cm}\)

For their magnetic moments to be equal:

\(M_A = M_B\)

This translates to:

\(N_A \cdot I_A \cdot \pi r_A^2 = N_B \cdot I_B \cdot \pi r_B^2\)

Simplifying gives:

\(N_A \cdot I_A \cdot r_A^2 = N_B \cdot I_B \cdot r_B^2\)

Substituting the given radii:

\(N_A \cdot I_A \cdot 10^2 = N_B \cdot I_B \cdot 20^2\)

Simplifying further, we have:

\(100 \cdot N_A \cdot I_A = 400 \cdot N_B \cdot I_B\)

Dividing both sides by 100:

\(N_A \cdot I_A = 4 \cdot N_B \cdot I_B\)

Thus, the condition for equal magnetic moments between circular coils A and B is satisfied by the equation:

\(N_A \cdot I_A = 4 \cdot N_B \cdot I_B\)

This confirms the correct answer is: Option 3: \(N _{ A } I _{ A }=4 N _{ B } I _{ B }\).