Q.
A straight wire is carrying a current of $100A$ as shown in the figure. What is the magnitude of the magnetic field at point $P$ which is at a perpendicular distance $\left(\right.\sqrt{3}-1\left.\right)m$ from the wire? (It is also given that ends $A$ and $B$ of the wire subtend angles $30^{^\circ }$ and $60^{^\circ }$ respectively at $P$ as shown)
NTA AbhyasNTA Abhyas 2020
Solution:
Magnetic field due to a straight current carrying conductor is given by
$B=\frac{\mu _{0} I}{4 \pi d}\left[\right.sin\alpha +sin\beta \left]\right.$
Here $\alpha $ is considered positive when measured clockwise and $\beta $ is considered positive when measured anticlockwise.
 <br/>
$\beta =150^\circ $ (anticlockwise) <br/>
Also, $\beta =-30^\circ $ (clockwise) <br/>
$d=\left(\sqrt{3} - 1\right)m$ and $i=100A$ <br/>
Plugging these values in the formula we get <br/>
$B=\frac{\left(10\right)^{- 7} \times 100}{\left(\sqrt{3} - 1\right)}\left[\frac{\sqrt{3}}{2} - \frac{1}{2}\right]=5\times \left(10\right)^{- 6}T$ <br/>
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Magnetic field due to a straight current carrying conductor is given by
$B=\frac{\mu _{0} I}{4 \pi d}\left[\right.sin\alpha +sin\beta \left]\right.$
Here $\alpha $ is considered positive when measured clockwise and $\beta $ is considered positive when measured anticlockwise.