Question

A coil having 9 turns carrying a current produces magnetic filed B₁ at the centre. Now the coil is rewounded into 3 turns carrying same current. Then the magnetic field at the centre B₂=

• A. \(\frac{B_1}{9}\)
• B. 9B₁
• C. 3B₁
• D. \(\frac{B_1}{3}\)

Answer 1

The Correct Option is A

Given:

• Initial coil has \( n_1 = 9 \) turns
• Produces magnetic field \( B_1 \) at the center
• New coil has \( n_2 = 3 \) turns, with same current and same wire (so same total length)

Step 1: Formula for Magnetic Field at Center of a Circular Coil

\[ B = \frac{\mu_0 n I}{2R} \] where:

• \( n \) = number of turns
• \( I \) = current
• \( R \) = radius of the coil

 

Step 2: Understand What Happens When Coil is Rewound

Since the total length of wire remains the same, fewer turns means each turn has a larger radius.

Let total length of wire be \( L \):

• Initial radius: \( R_1 \), \( L = 2\pi R_1 \cdot 9 \Rightarrow R_1 = \frac{L}{18\pi} \)
• New radius: \( R_2 \), \( L = 2\pi R_2 \cdot 3 \Rightarrow R_2 = \frac{L}{6\pi} \)

\[ \Rightarrow R_2 = 3R_1 \] Now apply the formula: \[ B_1 = \frac{\mu_0 \cdot 9I}{2R_1}, \quad B_2 = \frac{\mu_0 \cdot 3I}{2R_2} = \frac{\mu_0 \cdot 3I}{2 \cdot 3R_1} = \frac{\mu_0 \cdot I}{2R_1} \]

Step 3: Take the Ratio

\[ \frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2R_1}}{\frac{\mu_0 \cdot 9I}{2R_1}} = \frac{1}{9} \Rightarrow B_2 = \frac{B_1}{9} \]

The magnetic field at the center of the rewound coil is \( {\frac{B_1}{9}} \), so the correct answer is (A).

Answer 2

Let the initial number of turns be \( N_1 = 9 \). Let the radius of the coil be \( R_1 \). 

The magnetic field at the centre of this coil carrying current \( i \) is given by:

\[ B_1 = \frac{\mu_0 N_1 i}{2 R_1} = \frac{\mu_0 (9) i}{2 R_1} = \frac{9 \mu_0 i}{2 R_1} \]

Now, the coil is rewound into \( N_2 = 3 \) turns. Let the new radius be \( R_2 \).

Assuming the length of the wire used to make the coil remains constant:

Length \( L = N_1 \times (2 \pi R_1) = N_2 \times (2 \pi R_2) \)

\[ 9 \times (2 \pi R_1) = 3 \times (2 \pi R_2) \] \[ 9 R_1 = 3 R_2 \] \[ R_2 = 3 R_1 \]

The magnetic field at the centre of the new coil carrying the same current \( i \) is:

\[ B_2 = \frac{\mu_0 N_2 i}{2 R_2} \]

Substitute \( N_2 = 3 \) and \( R_2 = 3 R_1 \):

\[ B_2 = \frac{\mu_0 (3) i}{2 (3 R_1)} = \frac{3 \mu_0 i}{6 R_1} = \frac{\mu_0 i}{2 R_1} \]

Now, let's find the relationship between \( B_2 \) and \( B_1 \). We can find the ratio \( \frac{B_1}{B_2} \):

\[ \frac{B_1}{B_2} = \frac{\left( \frac{9 \mu_0 i}{2 R_1} \right)}{\left( \frac{\mu_0 i}{2 R_1} \right)} \] \[ \frac{B_1}{B_2} = 9 \] \[ B_1 = 9 B_2 \]

Therefore,

\[ B_2 = \frac{B_1}{9} \]

This corresponds to option (A).