Question

Let \( B_1 \) be the magnitude of magnetic field at the center of a circular coil of radius \( R \) carrying current \( I \). Let \( B_2 \) be the magnitude of magnetic field at an axial distance \( x \) from the center. For \( x : R = 3 : 4 \), \( \frac{B_2}{B_1} \) is:

• A. 4 : 5
• B. 16 : 25
• C. 64 : 125
• D. 25 : 16

Hint:

When dealing with the magnetic field produced by a circular coil, remember that the magnetic field is strongest at the center and weakens with distance along the axis.

Answer 1

The Correct Option is C

The magnetic field at the center of a circular coil is given by: \[ B_1 = \frac{\mu_0 I}{2R} \] The magnetic field at an axial distance \( x \) from the center is given by: \[ B_2 = B_1 \sin \theta = \frac{B_1 \cdot \left( \frac{R}{x} \right)^2}{5} \] Substituting \( x : R = 3 : 4 \), we get: \[ \frac{B_2}{B_1} = \frac{64}{125} \]

Answer 2

This problem asks for the ratio of the magnetic field at a point on the axis of a circular current-carrying coil to the magnetic field at its center. We are given the ratio of the axial distance to the radius of the coil.

Concept Used:

The solution requires the formulas for the magnetic field produced by a circular current-carrying coil at its center and at a point on its axis.

1. Magnetic Field at the Center (\( B_{\text{center}} \)): For a circular coil of radius \( R \) carrying a current \( I \), the magnitude of the magnetic field at its center is given by: \[ B_1 = B_{\text{center}} = \frac{\mu_0 I}{2R} \]
2. Magnetic Field on the Axis (\( B_{\text{axial}} \)): The magnitude of the magnetic field at a point on the axis of the coil, at a distance \( x \) from the center, is given by: \[ B_2 = B_{\text{axial}} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]

Here, \( \mu_0 \) is the permeability of free space.

Step-by-Step Solution:

Step 1: Formulate the ratio \( \frac{B_2}{B_1} \).

We need to find the ratio of the magnetic field on the axis (\( B_2 \)) to the magnetic field at the center (\( B_1 \)).

\[ \frac{B_2}{B_1} = \frac{\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}}{\frac{\mu_0 I}{2R}} \]

Step 2: Simplify the expression for the ratio.

We can cancel the common terms \( \frac{\mu_0 I}{2} \) from the numerator and the denominator:

\[ \frac{B_2}{B_1} = \frac{R^2}{(R^2 + x^2)^{3/2}} \times \frac{R}{1} \] \[ \frac{B_2}{B_1} = \frac{R^3}{(R^2 + x^2)^{3/2}} \]

Step 3: Use the given relationship between the axial distance \( x \) and the radius \( R \).

We are given that \( x : R = 3 : 4 \), which means:

\[ \frac{x}{R} = \frac{3}{4} \implies x = \frac{3}{4}R \]

Step 4: Substitute the value of \( x \) into the simplified ratio expression.

\[ \frac{B_2}{B_1} = \frac{R^3}{\left(R^2 + \left(\frac{3}{4}R\right)^2\right)^{3/2}} \] \[ \frac{B_2}{B_1} = \frac{R^3}{\left(R^2 + \frac{9}{16}R^2\right)^{3/2}} \] \[ \frac{B_2}{B_1} = \frac{R^3}{\left(\frac{16R^2 + 9R^2}{16}\right)^{3/2}} = \frac{R^3}{\left(\frac{25R^2}{16}\right)^{3/2}} \]

Final Computation & Result:

Now, we evaluate the denominator:

\[ \left(\frac{25R^2}{16}\right)^{3/2} = \left( \sqrt{\frac{25R^2}{16}} \right)^3 = \left( \frac{5R}{4} \right)^3 = \frac{125R^3}{64} \]

Substitute this back into the ratio:

\[ \frac{B_2}{B_1} = \frac{R^3}{\frac{125R^3}{64}} \]

Cancel the \( R^3 \) term:

\[ \frac{B_2}{B_1} = \frac{1}{\frac{125}{64}} = \frac{64}{125} \]

Therefore, the ratio \( \frac{B_2}{B_1} \) is \( \frac{64}{125} \).