Question

A wire is first bent in the form of a circular coil of 5 turns and the same wire is then bent in the form of another circular coil of 10 turns. If same current is passed in both the coils, then the ratio of the magnetic fields at their centres is

• A. 1:8
• B. 1:1
• C. 1:4
• D. 1:2

Hint:

When the total length of the wire is constant, the magnetic field strength is inversely proportional to the radius of the coil. Pay attention to the number of turns in each coil.

Answer 1

The Correct Option is C

To determine the ratio of the magnetic fields at the centers of two circular coils, we first need to understand the relationship between the magnetic field at the center of a circular coil and its parameters. The magnetic field \( B \) at the center of a single circular coil with \( N \) turns, carrying current \( I \), with radius \( R \) is given by the formula:

\( B = \frac{\mu_0 \cdot N \cdot I}{2 \cdot R} \)

where \( \mu_0 \) is the permeability of free space.

The problem states that there are two coils made from the same wire. The first coil has 5 turns, and the second coil has 10 turns. The current \( I \) is the same in both coils.

Since the same piece of wire is used, the total length of the wire for each coil remains constant. The length of the wire \( L \) can be represented for a single coil as:

\( L = N \cdot 2\pi R \)

For the first coil with 5 turns:

\( L = 5 \cdot 2 \pi R_1 \)

For the second coil with 10 turns:

\( L = 10 \cdot 2 \pi R_2 \)

Since both have the same wire length:

\( 5 \cdot 2 \pi R_1 = 10 \cdot 2 \pi R_2 \)

This implies:

\( R_1 = 2 \cdot R_2 \)

Now substituting these radii into the magnetic field formula, we find the magnetic fields for both coils.

The magnetic field for the first coil is:

\( B_1 = \frac{\mu_0 \cdot 5 \cdot I}{2 \cdot R_1} \)

The magnetic field for the second coil is:

\( B_2 = \frac{\mu_0 \cdot 10 \cdot I}{2 \cdot R_2} \)

Substituting \( R_1 = 2 \cdot R_2 \) into \( B_1 \):

\( B_1 = \frac{\mu_0 \cdot 5 \cdot I}{2 \cdot (2 \cdot R_2)} = \frac{\mu_0 \cdot 5 \cdot I}{4 \cdot R_2} \)

Now calculating the ratio of \( B_1 \) to \( B_2 \):

\( \frac{B_1}{B_2} = \frac{\frac{\mu_0 \cdot 5 \cdot I}{4 \cdot R_2}}{\frac{\mu_0 \cdot 10 \cdot I}{2 \cdot R_2}} \)

Simplifying:

\( \frac{B_1}{B_2} = \frac{5}{4} \cdot \frac{2}{10} = \frac{5}{20} = \frac{1}{4} \)

Answer 2

Step 1: Use the formula for magnetic field at the center of a circular coil.
The magnetic field at the center of a circular coil is given by:
\[ B = \frac{\mu_0 N I}{2R} \] where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space,
- \( N \) is the number of turns,
- \( I \) is the current,
- \( R \) is the radius of the coil.

Step 2: Understand the effect of bending the same wire into different coils.
Let the total length of the wire be \( L \).
For a coil of \( N \) turns, the total length of wire is distributed among \( N \) circles:
\[ L = N \cdot 2\pi R \Rightarrow R = \frac{L}{2\pi N} \]
So, radius \( R \) is inversely proportional to \( N \): \( R \propto \frac{1}{N} \)

Step 3: Substitute \( R \propto \frac{1}{N} \) into the magnetic field formula.
\[ B = \frac{\mu_0 N I}{2R} \propto \frac{N}{\frac{1}{N}} = N^2 \] Thus, magnetic field \( B \propto N^2 \)

Step 4: Take ratio of magnetic fields.
Let \( B_1 \) be the field for 5 turns and \( B_2 \) be the field for 10 turns:
\[ \frac{B_1}{B_2} = \left( \frac{5}{10} \right)^2 = \frac{25}{100} = \frac{1}{4} \]
So, the ratio \( B_1 : B_2 = 1 : 4 \)

Step 5: Conclusion.
The ratio of the magnetic fields at their centers is \( \boxed{1 : 4} \).