Question

Figure (a) shows a length of wire carrying a current i and bent into a circular coil of one turn. In Figure (b) the same length of wire has been bent to give a coil of 6 turns, each of 1/6 the original radius. (a) If B_a and B_b are the magnitudes of the magnetic fields at the centers of the two coils, what is the ratioB_b/B_a? (b) What is the ratio

Answer 1

Given that

Current through the loop is

Number of turns in first coil is \(n_{1}=1\)

Number of turns in second coil is \(n_{2}=3\) If \(R\) is the original radius of the coils, then radius of 6 turn coil is

\(r_{2}=\frac{R}{6}\)

From the ampheres law the magentic field ar ound the loop is

\(B=\frac{n \mu_{0} i}{2 \pi R}\)

The magnetic field of one turn coil is

$$ B_{a}=\frac{\mu_{0} i}{2 \pi R} $$

The magnetic field of six turn coil is

\(B_{a}=\frac{6 \mu_{0} i}{2 \pi r_{2}}\)

The ratio of two magnetic field is

\(\frac{B_{b}}{B_{a}}=\frac{\frac{6 \mu_{0} i}{2 \pi r_{2}}}{\frac{\mu_{0} i}{2 \pi R}}\)

$$ =\frac{\frac{6 \mu_{0} i}{2 \pi\left(\frac{R}{6}\right)}}{\frac{\mu_{0} i}{2 \pi R}} $$

\(=36\)

(b)

The ratio magnetic dipole moments is

$$ \begin{aligned} \frac{\mu_{b}}{\mu_{a}} &=\frac{6 r_{2}^{2}}{R^{2}} \\ &=\frac{6\left(\frac{R}{6}\right)^{2}}{R^{2}} \\ &=0.166 \end{aligned} $$