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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