Q.
A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b, respectively. When a current I passes through the coil, the magnetic field at the centre is
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a
μ0NIb
b
2μ0NIa
c
μ0NI2b−a ln ba
d
μ0IN2b−a ln ba
answer is C.
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Detailed Solution
Number of turns per unit width = Nb−a. Consider an elemental ring of radius x and with thickness dx Number of turns in the ring = dN= Ndxb−a. Magnetic field at the centre due to the ring elementdB=μ0dNi2x = μ0i2 Ndxb−a . 1x∴ Field at the centre∴ Field at the centre= ∫dB = μ0Ni2b−a ∫ab dxx=μ0Ni2b−a ln ba.
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