Question

If B₁ is the magnetic field induction at a point on the axis of a circular coil of radius R situated at a distance R✓3 and B2 is the magnetic field at the centre of the coil, then the ratio of is equal to K, find 16K. +

Answer 1

2

Answer 2

Solution:

1. Magnetic field at the center (B₂):

   $$B_2 = \frac{\mu_0 I}{2R}$$

2. Magnetic field on the axis at a distance $R\sqrt{3}$ (B₁):

   $$B_1 = \frac{\mu_0 I R^2}{2\left(R^2 + (R\sqrt{3})^2\right)^{3/2}}$$

   Simplify the denominator:

   $$R^2 + (R\sqrt{3})^2 = R^2 + 3R^2 = 4R^2$$

   Therefore,

   $$\left(4R^2\right)^{3/2} = 8R^3$$

   So,

   $$B_1 = \frac{\mu_0 I R^2}{2 \times 8 R^3} = \frac{\mu_0 I}{16R}$$

3. Ratio $K = \frac{B_1}{B_2}$:

   $$K = \frac{\frac{\mu_0 I}{16R}}{\frac{\mu_0 I}{2R}} = \frac{1}{16} \times \frac{2R}{R} = \frac{2}{16} = \frac{1}{8}$$

4. Finding $16K$:

   $$16K = 16 \times \frac{1}{8} = 2$$