A long, thin-walled pipe of radius R carries a current I along its length. The current density is uniform over the circumference of the pipe. The magnetic field at the centre of the pipe due to quarter portion of the pipe is
A
μ0I√24π2R
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B
μ0Iπ2R
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C
2μ0I√2π2R
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D
None
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Solution
The correct option is Aμ0I√24π2R
dI=I2πR×Rdθ=Idθ2π
Now, dB=μ0dI2πR=μ0Idθ4π2R →B=∫d→B=∫dBcosθ^i+∫dBsinθ^j=μ0I4π2R(∫cosθdθ^i+∫sinθdθ^j)
Integrating anf putting limits from 0 to π2, we get →B=μ0I4π2R(^i+^j)
Hence, B=μ0I√24π2R