Question: Q. 16. A circular coil, having 100 turns of wire, of radius (nearly) $20 \mathrm{~cm}$ each, lies in the $X Y$ plane with its centre at the Origin of co-ordinates. Find the magnetic field at the point $(0,0,20 \sqrt{3} \mathrm{~cm})$ when this coil carries a current of $\left(\frac{2}{\pi}\right) \mathrm{A}$.
A [Delhi Comptt. 2016]
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Solution:
Ans. The plane of coil is $X Y$ plane and field point is on the Z-axis.
$\therefore$ Magnetic field on the axial point
$$ \begin{array}{rlr} B & =\frac{\mu_{0} I R^{2} N}{2\left(R^{2}+z^{2}\right)^{3 / 2}} & \mathbf{1} \ & =\frac{4 \pi \times 10^{-7} \times \frac{2}{\pi} \times(0.2)^{2} \times 100}{2\left[(0.2)^{2}+(0.2 \sqrt{3})^{2}\right]^{3 / 2}} 1 / 2 \ & =\frac{8 \times 0.04 \times 10^{-7} \times 100}{2 \times 0.04 \times 8 \times 0.2} & 1 / 2 \ & =25 \mu \mathrm{T} & 1 / 2 \end{array} $$
[CBSE Marking Scheme 2016]
?Long Answer Type Questions
(5 marks each)
[AI Q. 1. (i) Derive an expression for the magnetic field at point on the axis of a current carrying circular loop.
(ii) A coil of 100 turns (tightly bound) and radius 10 $\mathrm{cm}$, carries a current of $1 \mathrm{~A}$. What is the magnitude of the magnetic field at the centre of the coil.
U] [SQP I 2017] Ans. (i) Derivation : 3 (ii) Numerical : 2 (i) Try yourself, Similar to 3 (b), Short Answer Type II
(ii) Since the coil is tightly bound, we may take each circular element to have the same radius $R=10 \mathrm{~cm}=0.1 \mathrm{~m}, \mathrm{~N}=100$ Magnitude of magnetic field :
$$ \begin{aligned} B & =\frac{\mu_{0} N I}{2 R} \ B & =\frac{4 \pi \times 10^{-7} \times 10^{2} \times 1}{2 \times 10^{-1}} \ & =2 \pi \times 10^{-4} \ & =6.28 \times 10^{-4} \mathrm{~T} \end{aligned} $$
[CBSE Marking Scheme 2017]
