Two concentric circular coils, each having 10 turns with radii $0.2 \mathrm{~m}$ and $0.4 \mathrm{~m}$ carry…
Two concentric circular coils, each having 10 turns with radii $0.2 \mathrm{~m}$ and $0.4 \mathrm{~m}$ carry currents $0.2 \mathrm{~A}$ and $0.3 \mathrm{~A}$ respectively in opposite directions. Magnetic field at the centre is
$(2 / 3) \mu_{0}$
$(5 / 4) \mu_{0}$
$(1 / 4) \mu_{0}$
$(1 / 6) \mu_{0}$
Solution
The magnetic field at the centre of a circular coil carrying current $I$ is given by
$B=\frac{\mu_{0} n I}{2 R}$
where, $n$ = number of turns.
For coil $1, B_{1}=\frac{\mu_{0} \times 10 \times 0.2}{2 \times 0.2}=5 \mu_{0}$
For coil 2, $B_{2}=\frac{\mu_{0} \times 10 \times 0.3}{2 \times 0.4}=\frac{15}{4} \mu_{0}$
$\therefore$ Net magnetic field,
$B=B_{1}-B_{2}=5 \mu_{0}-\frac{15}{4} \mu_{0}=\frac{5}{4} \mu_{0}$