Magnetic Effects & Magnetism

JEE Main level, single correct, medium

46 of 276

Question 46 of 276

A long conducting wire having a current II flowing through it, is bent into a circular coil of NN turns. Then it is bent into a circular coil of nn 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:

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Answer: (C) N2:n2{N}^{2}:{n}^{2}

As in both cases, length of the wire will remain the same, therefore
l=(2πR)N=(2πr)n⇒Rr=nNl=\left(2\pi R\right)N=\left(2\pi r\right)n \Rightarrow \dfrac{R}{r}=\dfrac{n}{N}
Now magnetic field due to single circular coil is given by, B=(μ0I2r)B=\left(\dfrac{{\mu }_{0}I}{2r}\right).
Therefore, required ratio of total magnetic field wil be,
B1B2=N(μ0I2R)n(μ0I2r)=(N2n2)\dfrac{{B}_{1}}{{B}_{2}}=\dfrac{N\left(\dfrac{{\mu }_{0}I}{2R}\right)}{n\left(\dfrac{{\mu }_{0}I}{2r}\right)}=\left(\dfrac{{N}^{2}}{{n}^{2}}\right)

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