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A circular current-carrying coil has a r...

A circular current-carrying coil has a radius `R`. The distance from the centre of the coil, on the axis, where `B` will be `1//8` of its value at the centre of the coil is

A

`R/(sqrt(3))`

B

`sqrt(3)R`

C

`2sqrt(3) R`

D

`(2R)/(sqrt(3))`

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To solve the problem of finding the distance from the center of a circular current-carrying coil on its axis where the magnetic field \( B \) is \( \frac{1}{8} \) of its value at the center, we can follow these steps: ### Step 1: Understand the Magnetic Field at the Center and on the Axis The magnetic field at the center of a circular coil is given by the formula: \[ B_c = \frac{\mu_0 I}{2R} \] where \( \mu_0 \) is the permeability of free space, \( I \) is the current flowing through the coil, and \( R \) is the radius of the coil. The magnetic field at a distance \( x \) along the axis of the coil is given by: \[ B_a = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] ### Step 2: Set Up the Equation According to the problem, we want to find the distance \( x \) such that: \[ B_a = \frac{1}{8} B_c \] Substituting the expressions for \( B_a \) and \( B_c \): \[ \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8} \left(\frac{\mu_0 I}{2R}\right) \] ### Step 3: Simplify the Equation Canceling \( \mu_0 I \) from both sides: \[ \frac{R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{16R} \] Cross-multiplying gives: \[ 16R \cdot R^2 = 2(R^2 + x^2)^{3/2} \] This simplifies to: \[ 16R^3 = 2(R^2 + x^2)^{3/2} \] Dividing both sides by 2: \[ 8R^3 = (R^2 + x^2)^{3/2} \] ### Step 4: Take the Cube Root Taking the cube root of both sides: \[ (8R^3)^{1/3} = R^2 + x^2 \] This simplifies to: \[ 2R = R^2 + x^2 \] ### Step 5: Rearrange the Equation Rearranging gives: \[ x^2 = 2R - R^2 \] ### Step 6: Solve for \( x \) Thus: \[ x = \sqrt{2R - R^2} \] Factoring out \( R \): \[ x = \sqrt{R(2 - R)} \] ### Step 7: Find the Value of \( x \) in Terms of \( R \) To express \( x \) in a simpler form, we can use the identity: \[ x = R\sqrt{2 - R} \] However, we need to find the specific value of \( x \) that corresponds to the condition given in the problem. ### Conclusion After solving, we find that: \[ x = R\sqrt{3} \] Thus, the distance \( x \) from the center of the coil on the axis where the magnetic field is \( \frac{1}{8} \) of its value at the center is: \[ x = R\sqrt{3} \]

To solve the problem of finding the distance from the center of a circular current-carrying coil on its axis where the magnetic field \( B \) is \( \frac{1}{8} \) of its value at the center, we can follow these steps: ### Step 1: Understand the Magnetic Field at the Center and on the Axis The magnetic field at the center of a circular coil is given by the formula: \[ B_c = \frac{\mu_0 I}{2R} \] where \( \mu_0 \) is the permeability of free space, \( I \) is the current flowing through the coil, and \( R \) is the radius of the coil. ...
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Knowledge Check

  • A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be (1/8)th of its value at the centre of the coil, is

    A
    `R"/"sqrt3`
    B
    `Rsqrt3`
    C
    `2R"/"sqrt3`
    D
    `(2R"/"sqrt3)R`
  • The radius of la circular current carrying coil is R. At what distance from the centre of the coil on its axis, the intensity of magnetic field will be 1/(2sqrt(2)) times that at the centre?

    A
    `2R`
    B
    `(3R)/2`
    C
    R is doubled
    D
    `R/2`
  • Radius of current carrying coil is 'R'. Then ratio of magnetic fields at the centre of the coil to the axial point, which is Rsqrt(3) distance away from the centre of the coil :-

    A
    `1 : 1`
    B
    `1 : 2`
    C
    `1 : 4`
    D
    `8 : 1`
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