Q150 / 160 · AP EAMCET 2023 (18 May Shift 1) · Physics

Consider a tightly wound 100 turn coil of radius $10 \mathrm{~cm}$ carrying a current of $2 \mathrm{~A}$. The magnitude of the magnetic field at the centre of the coil is

  • A
    $3.14 \times 10^{-4} \mathrm{~T}$
  • B
    $6.28 \times 10^{-4} \mathrm{~T}$
  • Correct answer
    C
    $12.56 \times 10^{-4} \mathrm{~T}$
  • D
    $0$

Solution

Number of turns $n=100$ Radius, $r=10 \mathrm{~cm}=0.1 \mathrm{~m}$ current, $\mathrm{I}=2 \mathrm{~A}$ Magnetic field at center of coil: $\begin{aligned} & \mathrm{B}=\frac{\mu_0 \mathrm{ni}}{2 \mathrm{r}}=\frac{4 \pi \times 100 \times 2 \times 10}{2 \times 0.1} \times 10^{-7} \\ & \mathrm{~B}=12.56 \times 10^{-4} \mathrm{~T}\end{aligned}$

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

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