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 $200\, mA$ 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 )
$N_{1}=150, R_{1}=\pi\, cm =\pi \times 10^{-2} \,m$
$I_{1}=200 \,mA =200 \times 10^{-3} \,A$
For coil 2 (In vertical plane)
$N_{2}=400, R_{2}=2 \pi \,cm =2 \pi \times 10^{-2} \,m$
$I_{2}=200 \,mA =200 \times 10^{-3}\, A$
Magnetic field at the common centre $O$ due to current in coil 1 is
$B_{1}=\frac{\mu_{0}}{4 \pi} \frac{N_{1} I_{1} 2 \pi}{R_{1}}=\frac{\mu_{0} N_{1} I_{1}}{2 R_{1}}$
Magnetic field at $O$ due to current in coil 2 is
$B_{2}=\frac{\mu_{0}}{4 \pi} \frac{N_{2} I_{2} 2 \pi}{R_{2}}=\frac{\mu_{0} N_{2} I_{2}}{2 R_{2}}$
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
$B=\sqrt{\left(\frac{\mu_{0} \times 150 \times 200 \times 10^{-3}}{2 \times \pi \times 10^{-2}}\right)^{2}+\left(\frac{\mu_{0} \times 400 \times 200 \times 10^{-3}}{2 \times 2 \pi \times 10^{-2}}\right)^{2}}$
$=\frac{\mu_{0} \times 200 \times 10^{-3} \times 10}{2 \times \pi \times 10^{-2}} \sqrt{(15)^{2}+(20)^{2}}$
$=10^{-3} \,Wb \,m ^{-2}$