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# Stuck here, help me understand: A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B It is then bent into a circular loop of n turns. The magnetic field at the centre of th

A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B It is then bent into a circular loop of  n turns. The magnetic field at the centre of the coil will be

• Option 1)

$nB$

• Option 2)

$n^{2}B$

• Option 3)

$2nB$

• Option 4)

$2n^{2}B$

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As we discussed in

Magnetic Field due to Circular Current at the centre -

If a coil of radius r is carrying current i then magnetic feild at a distnace X from its centre is:

$B_{centre}= \frac{\mu_{0} }{4\pi } \frac{2\pi Ni}{r} = \frac{\mu_{0} Ni}{2r}$

- wherein

$:\; \; Initially,\; \; r_{1}= \: radius \: of \: coil \: = \frac{l}{2}\pi$

$\therefore \: \: \: B= \frac{\mu _{0}i}{2r_{i}}= \frac{2 \mu _{o}i\pi}{2l}$

$Finally\: \; \; r_{2}= radius\: of \: coil= \frac{l}{2\pi n}$

$\therefore \: \: \; \; {B}'=\frac{\mu _{0}i\times n}{2r_{2}}= \frac{n\mu_{0}i\times 2\pi n }{2l}= \frac{2\mu _{0}in^{2}\pi }{2l}$

$\frac{{B}'}{B}= \frac{2\mu _{0}in^{2}\pi }{2l}\times \frac{2l}{2\mu _{0}i\pi }= n^{2}$

${B}'= n^{2}B$

Option 1)

$nB$

Incorrect option

Option 2)

$n^{2}B$

Correct option

Option 3)

$2nB$

Incorrect option

Option 4)

$2n^{2}B$

Incorrect option

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