A tightly wound coil of 200 turns and of radius 20 cm carrying current 5 A . Magnetic field at the centre of…

A tightly wound coil of 200 turns and of radius 20 cm carrying current 5 A . Magnetic field at the centre of the coil is
  1. $3.14 \times 10^{-3} \mathrm{~T}$
  2. $3.14 \times 10^{-2} \mathrm{~T}$
  3. $6.28 \times 10^{-4} \mathrm{~T}$
  4. $6.28 \times 10^{-3} \mathrm{~T}$

Solution

For tightly wound coil, $\mathrm{N}=200, \mathrm{r}=20 \mathrm{~cm}, \mathrm{I}=5 \mathrm{~A}$ $\therefore$ Magnetic field at the centre of coil, $\begin{aligned} & \mathrm{B}=\frac{\mu_0 \mathrm{IN}}{2 \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 5 \times 200}{2 \times 20 \times 10^{-2}} \\ & =3.14 \times 10^{-3} \mathrm{~T} \end{aligned}$

Asked in: AP EAMCET 2024 (19 May Shift 2)

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