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  4. A straight current element is carrying current i=100 A, as shown in the figure. The magnitude of magnetic field at point P which is at perpendicular distance (√3-1) m from the current element, if end A and end B of the element subtends angle 30° and 60° at point P, respectively, as shown, is n × 10-6 tesla. Find the value of n.

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Q. A straight current element is carrying current $i=100 A$, as shown in the figure. The magnitude of magnetic field at point $P$ which is at perpendicular distance $(\sqrt{3}-1) m$ from the current element, if end $A$ and end $B$ of the element subtends angle $30^{\circ}$ and $60^{\circ}$ at point $P$, respectively, as shown, is $n \times 10^{-6}$ tesla. Find the value of $n$.Physics Question Image

Moving Charges and Magnetism

Solution:

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$ B =\frac{\mu_{0} . i}{4 \pi d}\left(\sin \theta_{1}+\sin \theta_{2}\right)$
$=\frac{10^{-7} \times 100}{\sqrt{3}-1}\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\right] $
$=5 \times 10^{-6} T $