Question

Two flat circular coils have a common centre, but their planes are at right angles to each other. The inner coil has 150 turns and radius of $\pi cm$. The outer coil has 400 turns and a radius of $2\pi cm$. The magnitude of the resultant magnetic field at the common centres of the coils wher a current of $200mA$ is sent through each of them is

• A. $10^{-3}Wb{m}^{-2}$
• B. $2\times 10^{-3}Wb{m}^{-2}$
• C. $5\times 10^{-3}Wb{m}^{-2}$
• D. $7\times 10^{-3}Wb{m}^{-2}$

Answer 1

Answer: (A)

For coil 1 (In horizontal plane )
$N_1=150,R_1=\pi cm=\pi \times 10^{-2}m$
$I_1=200mA=200\times 10^{-3}A$
For coil 2 (In vertical plane)
$N_2=400,R_2=2\pi cm=2\pi \times 10^{-2}m$
$I_2=200mA=200\times 10^{-3}A$
Magnetic field at the common centre $O$ due to current in coil 1 is
$B_1=\frac{{\mu }_0}{4\pi }\frac{N_1{I}_{1}2\pi }{R_1}=\frac{{\mu }_0{N}_1{I}_1}{2R_1}$
Magnetic field at $O$ due to current in coil 2 is
$B_2=\frac{{\mu }_0}{4\pi }\frac{N_2{I}_{2}2\pi }{R_2}=\frac{{\mu }_0{N}_2{I}_2}{2R_2}$
Resultant magnetic field at $O$ is
$B=\sqrt{B_1^2+B_2^2}$
$(\because B_1$ is $\perp$ to $B_2)$
Substituting the given values, we get
$B=\sqrt{{\left(\frac{{\mu }_0\times 150\times 200\times 10^{-3}}{2\times \pi \times 10^{-2}}\right)}^2+{\left(\frac{{\mu }_0\times 400\times 200\times 10^{-3}}{2\times 2\pi \times 10^{-2}}\right)}^2}$
$=\frac{{\mu }_0\times 200\times 10^{-3}\times 10}{2\times \pi \times 10^{-2}}\sqrt{(15{)}^2+(20{)}^2}$
$=10^{-3}Wb{m}^{-2}$