Question

The magnetic field due to a current carrying circular coil on its axis at a distance of \(\sqrt{2} d\) from the centre of the coil is B. If d is the diameter of the coil, then the magnetic field at the centre of the coil is

• A. 18B
• B. 27B
• C. 3B
• D. 9B

Hint:

When comparing magnetic fields at two different points for the same coil, it's best to write out the formulas for both and then take their ratio, or solve for the constant term (\(\mu_0 I / r\)) as done here. Be careful with algebra, especially with exponents like \( (r^2)^{3/2} = r^3 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves the formulas for the magnetic field produced by a circular current loop, both at its center and at a point along its axis. We need to establish a relationship between the two.
Step 2: Key Formula or Approach:
Let \(r\) be the radius of the coil.
- The magnetic field at the center of the coil is: \(B_{center} = \frac{\mu_0 I}{2r}\).
- The magnetic field at a distance \(x\) from the center on its axis is: \(B_{axis} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}\).
Step 3: Detailed Explanation:
First, let's relate the given distances to the radius \(r\).
- Diameter of the coil = \(d\).
- Radius of the coil, \(r = d/2\).
- The distance from the center, \(x = \sqrt{2} d = \sqrt{2} (2r) = 2\sqrt{2} r\).
We are given that the field at this distance \(x\) is \(B\). Let's write the expression for \(B\):
\[ B = B_{axis} = \frac{\mu_0 I r^2}{2(r^2 + (2\sqrt{2} r)^2)^{3/2}} \] Now, simplify the denominator:
\[ r^2 + (2\sqrt{2} r)^2 = r^2 + 8r^2 = 9r^2 \] So, the expression for B becomes:
\[ B = \frac{\mu_0 I r^2}{2(9r^2)^{3/2}} = \frac{\mu_0 I r^2}{2 \times 9^{3/2} \times (r^2)^{3/2}} = \frac{\mu_0 I r^2}{2 \times (3^2)^{3/2} \times r^3} = \frac{\mu_0 I r^2}{2 \times 3^3 \times r^3} = \frac{\mu_0 I r^2}{2 \times 27 \times r^3} \] \[ B = \frac{\mu_0 I}{54r} \quad \cdots(1) \] We need to find the magnetic field at the center, \(B_{center}\):
\[ B_{center} = \frac{\mu_0 I}{2r} \quad \cdots(2) \] To relate \(B_{center}\) to \(B\), we can express \(\frac{\mu_0 I}{r}\) from equation (1):
\[ \frac{\mu_0 I}{r} = 54B \] Now substitute this into equation (2):
\[ B_{center} = \frac{1}{2} \left( \frac{\mu_0 I}{r} \right) = \frac{1}{2} (54B) = 27B \] Step 4: Final Answer:
The magnetic field at the centre of the coil is 27B. Therefore, option (B) is correct.