Q128 / 170 · Manipal (MET) 2016 · Physics

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

  • A
    $n B$
  • Correct answer
    B
    $n^2 B$
  • C
    $2 n B$
  • D
    $2 n^2 B$

Solution

$\therefore$ Current, $l=2 \pi R=2 \pi n \cdot r$ $\left [\because r=\frac{R}{n}\right]$ The magnetic field at the centre of the coil, $B=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi I}{R}$\ldots(i) and $B^{\prime}=\left(\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi I}{r}\right) \cdot n$\ldots(ii) From Eqs. (i) and (ii), we get $B^{\prime}=n^2 B$

Hence, the correct answer is option: 'B'.

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