Question:
A coil having 9 turns carrying current produces magnetic field \( B_1 \) at the centre. Now that coil is rewound into 3 turns carrying same current. Then magnetic field at the centre \( B_2 \) is
A coil having 9 turns carrying current produces magnetic field \( B_1 \) at the centre. Now that coil is rewound into 3 turns carrying same current. Then magnetic field at the centre \( B_2 \) is
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The magnetic field produced by a coil is directly proportional to the number of turns. If the number of turns is reduced, the magnetic field decreases by the same factor.
Updated On: Jun 30, 2026
- \( \frac{B_1}{9} \)
- \( 9B_1 \)
- \( 3B_1 \)
- \( \frac{B_1}{3} \)
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The Correct Option is D
Solution and Explanation
Step 1: Formula for magnetic field due to a coil.
The magnetic field at the center of a coil carrying current is given by: \[ B = \frac{\mu_0 N I}{2R}, \] where: - \( N \) is the number of turns in the coil, - \( I \) is the current, - \( R \) is the radius of the coil, - \( \mu_0 \) is the permeability of free space.
Step 2: Magnetic field for the original coil.
For the original coil with \( N = 9 \) turns, the magnetic field at the center is: \[ B_1 = \frac{\mu_0 \cdot 9 I}{2R}. \]
Step 3: Magnetic field for the new coil.
When the coil is rewound into 3 turns, the magnetic field at the center becomes: \[ B_2 = \frac{\mu_0 \cdot 3 I}{2R}. \]
Step 4: Finding the ratio of the magnetic fields.
The ratio of the magnetic field \( B_2 \) to \( B_1 \) is: \[ \frac{B_2}{B_1} = \frac{\frac{\mu_0 \cdot 3 I}{2R}}{\frac{\mu_0 \cdot 9 I}{2R}} = \frac{3}{9} = \frac{1}{3}. \] Final Answer:
Thus, the magnetic field at the center for the coil with 3 turns is: \[ \boxed{\frac{B_1}{3}}. \]
The magnetic field at the center of a coil carrying current is given by: \[ B = \frac{\mu_0 N I}{2R}, \] where: - \( N \) is the number of turns in the coil, - \( I \) is the current, - \( R \) is the radius of the coil, - \( \mu_0 \) is the permeability of free space.
Step 2: Magnetic field for the original coil.
For the original coil with \( N = 9 \) turns, the magnetic field at the center is: \[ B_1 = \frac{\mu_0 \cdot 9 I}{2R}. \]
Step 3: Magnetic field for the new coil.
When the coil is rewound into 3 turns, the magnetic field at the center becomes: \[ B_2 = \frac{\mu_0 \cdot 3 I}{2R}. \]
Step 4: Finding the ratio of the magnetic fields.
The ratio of the magnetic field \( B_2 \) to \( B_1 \) is: \[ \frac{B_2}{B_1} = \frac{\frac{\mu_0 \cdot 3 I}{2R}}{\frac{\mu_0 \cdot 9 I}{2R}} = \frac{3}{9} = \frac{1}{3}. \] Final Answer:
Thus, the magnetic field at the center for the coil with 3 turns is: \[ \boxed{\frac{B_1}{3}}. \]
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