Verified Solution • 2025–26 Session moderate

QJEEM23Pha0b9: JEE Main 2023 Physics - Class 12 JEEM23Pha0b9 Solution: A long conducting wire having a current I flowing through it,…

A long conducting wire having a current I flowing through it, is bent into a circular coil of N\mathrm{N} turns. Then it is bent into a circular coil of n\mathrm{n} turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

Multiple Choice Options

  • A N2:n2N^{2}: n^{2}
  • B N:n\mathrm{N}: \mathrm{n}
  • C n:N\mathrm{n}: \mathrm{N}
  • D n2:N2n^{2}: N^{2}

Step-by-Step Verified Solution

I=(2πr)nI=(2 \pi r) n

amp;r∝(In)amp;B=n(μ0i2r)∝(μ0i2L)n2amp;B1B2=(N2n2)\begin{aligned}& r \propto\left(\frac{I}{n}\right) \\\\& B=n\left(\frac{\mu_{0} i}{2 r}\right) \propto\left(\frac{\mu_{0} i}{2 L}\right) n^{2} \\\\& \frac{B_{1}}{B_{2}}=\left(\frac{N^{2}}{n^{2}}\right)\end{aligned}

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