KCET · Physics · Electrostatics
Two concentric coils each of radius equal to \( 2 \Pi \mathrm{cm} \) are placed right angles to each other. If \( 3 \mathrm{~A} \) and \( 4 \mathrm{~A} \) are the currents flowing through the two coils respectively. The magnetic induction (in \( \mathrm{Wb} \mathrm{m}^{-2} \) ) at the centre of the coils will be
- A \( 12 \times 10^{-5} \)
- B \( 10^{-5} \)
- C \( 5 \times 10^{-5} \)
- D \( 7 \times 10^{-5} \)
Answer & Solution
Correct Answer
(C) \( 5 \times 10^{-5} \)
Step-by-step Solution
Detailed explanation
Magnetic induction at the centre of coil is given as
\(B=\frac{\mu_{0}}{2} \frac{I}{a}\)
where \( \mathrm{a} \) is radius of coil.
Here, \( a=2 \pi \mathrm{cm}=2 \Pi \times 10^{-2} \mathrm{~m} \)
For current
\(I_{1}=3 A, B_{1}=\frac{\mu_{0}}{2} \frac{3}{2 \pi \times 10^{-2}}=3 \times 10^{-5} T\)
For current
\(I_{2}=4 A, B_{2}=\frac{\mu_{0}}{2} \frac{4}{2 \Pi \times 10^{-2}}=4 \times 10^{-5} \mathrm{~T}\)
Then, magnetic induction at centre of coils is
\(B=\sqrt{B_{1}{ }^{2}+B_{2}{ }^{2}}=\sqrt{\left(3 \times 10^{-5}\right)^{2}+\left(4 \times 10^{-5}\right)^{2}}=5 \times 10^{-5} \mathrm{~T}\)
\(B=\frac{\mu_{0}}{2} \frac{I}{a}\)
where \( \mathrm{a} \) is radius of coil.
Here, \( a=2 \pi \mathrm{cm}=2 \Pi \times 10^{-2} \mathrm{~m} \)
For current
\(I_{1}=3 A, B_{1}=\frac{\mu_{0}}{2} \frac{3}{2 \pi \times 10^{-2}}=3 \times 10^{-5} T\)
For current
\(I_{2}=4 A, B_{2}=\frac{\mu_{0}}{2} \frac{4}{2 \Pi \times 10^{-2}}=4 \times 10^{-5} \mathrm{~T}\)
Then, magnetic induction at centre of coils is
\(B=\sqrt{B_{1}{ }^{2}+B_{2}{ }^{2}}=\sqrt{\left(3 \times 10^{-5}\right)^{2}+\left(4 \times 10^{-5}\right)^{2}}=5 \times 10^{-5} \mathrm{~T}\)
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