Question

Two concentric circular coils, each having 10 turns with radii 0.2 m and 0.4 m carry currents 0.2 A and 0.3 A respectively in opposite directions. Magnetic field at the centre is

• A. $(2 / 3) \, \mu_0$
• B. $(5 / 4) \, \mu_0$
• C. $(1 / 4) \, \mu_0$
• D. $(1 / 6) \, \mu_0$

Answer 1

Answer: (B)

$(5 / 4) \, \mu_0$

Answer 2

Here, $r_1 = 0.2 \, m$, $r_2 = 0.4 \, m$, $I_1 = 0.2 \, A$, $I_2 = 0.3 \, A$ $n_1 = n_2 = n = 10$ Magnetic field due to smaller loop at the centre $O$, $\bar{B}_1 = \frac{ n \mu_0 I_1}{2 r_1}\bigotimes$ $= \frac{10 \times \mu_0 \times 0.2}{2 \times 0.2} = 5 \mu \bigotimes$ Magnetic field due to larger loop at centre $O$, $\bar{B}_2 = \frac{ n \mu_0 I_2}{2 r_1} \bigodot$ $= \frac{10 \times \mu_0 \times 0.3}{2 \times 0.4} = \frac{15}{4} \mu \bigodot$ Net magnetic field at the centre, $\bar{B}_0 = \bar{B}_1 - \bar{B}_2$ $= 5 \mu_0 - \frac{15}{4} \mu_0 = \frac{5 \mu_0}{4}$