Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

4 of 276

Question 4 of 276

The ratio of magnetic field at the centre of a current carrying coil of radius rr to the magnetic field at distance rr from the centre of coil on its axis is x:1\sqrt{x}:1. The value of xx is _____.

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Answer: 8

The formula to calculate the magnetic field at any axial point situated at a distance xx from the centre of a circular current carrying conductor of radius rr can be written as
B=μ0Ir22(r2+x2)3/2...(1)B=\dfrac{{\mu }_{0}I{r}^{2}}{2{\left({r}^{2}+{x}^{2}\right)}^{3/2}}...\left(1\right)
Substitute 00 for xx in equation (1) to calculate the magnetic field (Bc)\left({B}_{c}\right) at the centre of the conductor.
Bc=μ0Ir22(r2+02)3/2=μ0Ir22r3=μ0I2r...(2)\begin{matrix}{B}_{c} & = & \dfrac{{\mu }_{0}I{r}^{2}}{2{\left({r}^{2}+{0}^{2}\right)}^{3/2}} \\ & = & \dfrac{{\mu }_{0}I{r}^{2}}{2{r}^{3}} \\ & = & \dfrac{{\mu }_{0}I}{2r}...\left(2\right)\end{matrix}
Substitute rr for xx in equation (1) to calculate the magnetic field (Br)\left({B}_{r}\right) at the given distance from the centre of the conductor.
Br=μ0Ir22(r2+r2)3/2=μ0Ir242r3=μ0I42r...(3)\begin{matrix}{B}_{r} & = & \dfrac{{\mu }_{0}I{r}^{2}}{2{\left({r}^{2}+{r}^{2}\right)}^{3/2}} \\ & = & \dfrac{{\mu }_{0}I{r}^{2}}{4\sqrt{2}{r}^{3}} \\ & = & \dfrac{{\mu }_{0}I}{4\sqrt{2}r}...\left(3\right)\end{matrix}
Divide equation (2) by equation (3) to calculate the required ratio of the magnetic fields.
BcBr=μ0I2rμ0I42r=22=8\begin{matrix}\dfrac{{B}_{c}}{{B}_{r}} & = & \dfrac{\dfrac{{\mu }_{0}I}{2r}}{\dfrac{{\mu }_{0}I}{4\sqrt{2}r}} \\ & = & 2\sqrt{2} \\ & = & \sqrt{8}\end{matrix}

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