Question

At what distance x from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?

Answer 1

Answer:

The value of x is 2.1 cm from the center of the coil.

Explanation:

Radius, R = 2.7 cm

Number of turns, N = 800

The magnetic field at the axis is half of the magnetic field at the center.

$B_{axis}=\frac{B_{center}}{2}\\\\\frac{\mu o}{4\pi}\times \frac{2 \pi I N R^2}{\left (R^2 + x^2 \right )^{\frac{3}{2}}} = 0.5\frac{\mu o}{4\pi}\times\frac{2\pi N I}{R}\\\\\frac{R^2}{(R^2 + x^2)^\frac{3}{2}} = \frac{1}{2R}\\\\4R^6 = (R^2+x^2)^3\\\\1.6 R^2 = R^2 + x^2\\\\x^2 = 0.6 \times 2.7\times 2.7 \\\\x = 2.1 cm$