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Numericals · Q19

Q.A solenoid of length π\pi m and 5 cm in diameter has a winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the axis.

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The turns per unit length is n=N/ℓ=1000/πn=N/\ell=1000/\pi (length ℓ=π\ell=\pi m). Using the ideal solenoid formula (Section 10.16.1), B=μ0ni=(4π×10−7)(1000π)(5)B=\mu_0ni=(4\pi\times10^{-7})\left(\dfrac{1000}{\pi}\right)(5). The π\pi in the numerator and denominator cancel: B=(4×10−7)(1000)(5)=2×10−3B=(4\times10^{-7})(1000)(5)=2\times10^{-3} T. Note that …

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