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A wire of certain length carries a curre...

A wire of certain length carries a current l.It is bent to form a circle of one turn and the magnetic field at the centre is B1. If it is bent to form a coil of four turns, then magnetic field at centre is `B_2`.The ratio of `B_1` and `B_2 `is

A

`1 : 16 `

B

`1 : 4`

C

`64 : 1 `

D

`2 : 1 `

Text Solution

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The correct Answer is:
To solve the problem, we need to determine the ratio of the magnetic fields \( B_1 \) and \( B_2 \) produced at the center of a circular wire when it is bent into one turn and then into four turns. ### Step-by-Step Solution: 1. **Understanding the Magnetic Field Formula**: The magnetic field at the center of a circular loop carrying current \( I \) is given by the formula: \[ B = \frac{\mu_0 n I}{2R} \] where: - \( B \) is the magnetic field, - \( \mu_0 \) is the permeability of free space, - \( n \) is the number of turns, - \( I \) is the current, and - \( R \) is the radius of the loop. 2. **Calculating \( B_1 \)**: For the first case, where the wire is bent into one turn: - Number of turns \( n_1 = 1 \) - Let the radius of the loop be \( R_1 = R \). Thus, the magnetic field \( B_1 \) is: \[ B_1 = \frac{\mu_0 \cdot 1 \cdot I}{2R} = \frac{\mu_0 I}{2R} \] 3. **Calculating \( B_2 \)**: For the second case, where the wire is bent into four turns: - Number of turns \( n_2 = 4 \) - The length of the wire remains the same, so the radius of the new loop \( R_2 \) is related to \( R \) by the inverse of the number of turns: \[ R_2 = \frac{R}{4} \] Thus, the magnetic field \( B_2 \) is: \[ B_2 = \frac{\mu_0 \cdot 4 \cdot I}{2R_2} = \frac{\mu_0 \cdot 4 \cdot I}{2 \cdot \frac{R}{4}} = \frac{\mu_0 \cdot 4 \cdot I \cdot 4}{2R} = \frac{16 \mu_0 I}{2R} = \frac{8 \mu_0 I}{R} \] 4. **Finding the Ratio \( \frac{B_1}{B_2} \)**: Now we can find the ratio of \( B_1 \) to \( B_2 \): \[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2R}}{\frac{8 \mu_0 I}{R}} = \frac{\mu_0 I}{2R} \cdot \frac{R}{8 \mu_0 I} = \frac{1}{16} \] Thus, the ratio of \( B_1 \) to \( B_2 \) is: \[ \frac{B_1}{B_2} = \frac{1}{16} \] ### Final Answer: The ratio of \( B_1 \) and \( B_2 \) is \( \frac{1}{16} \). ---
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Knowledge Check

  • A wire of length L carrying a current I is bent into a circle. The magnitude of the magnetic field at the centre of the circle is

    A
    `(pi mu_(0)I)/(L)`
    B
    `(mu_(0)I)/(2L)`
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    `(2pi mu_(0)I)/L`
    D
    `(mu_(0)I)/(2 pi L)`
  • A wire of length L carrying a current I is bent into a circle. The magnitude of the magneitc field at the centre of the circle is

    A
    `(pi mu_(0)I)/(L)`
    B
    `(mu_(0)I)/(2L)`
    C
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  • A long wire carries a steady curent . It is bent into a circle of one turn and the magnetic field at the centre of the coil is B . It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be

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    ` 2n B`
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