Find the magnetic field at point P in the giv | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The magnetic field due to a finite straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4 \pi r}(\sin 	heta_1 + \sin 	h

Vedclass · Accepted Answer

$\frac{\mu_0 i}{8 \pi a}(\sqrt{3}-1) \odot$

Vedclass · Answer

(C) The magnetic field due to a finite straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4 \pi r}(\sin 	heta_1 + \sin 	heta_2)$.In the given diagram,the perpendicular distance from point $P$ to the wire is $r = a \cos 30^\circ = a \frac{\sqrt{3}}{2}$.The angles subtended by the ends of the wire at point $P$ are $	heta_1 = 30^\circ$ and $	heta_2 = 60^\circ$.Substituting these values into the formula:$B = \frac{\mu_0 i}{4 \pi (a \frac{\sqrt{3}}{2})} (\sin 60^\circ - \sin 30^\circ)$Note: Since the wire segment is on one side of the perpendicular,we subtract the angles.$B = \frac{\mu_0 i}{2 \pi a \sqrt{3}} (\frac{\sqrt{3}}{2} - \frac{1}{2}) = \frac{\mu_0 i}{2 \pi a \sqrt{3}} (\frac{\sqrt{3}-1}{2}) = \frac{\mu_0 i}{4 \pi a} (1 - \frac{1}{\sqrt{3}})$.However,re-evaluating the geometry: the perpendicular distance is $a \cos 30^\circ$. The field is $B = \frac{\mu_0 i}{4 \pi r}(\sin 	heta_2 - \sin 	heta_1) = \frac{\mu_0 i}{4 \pi (a \cos 30^\circ)} (\sin 60^\circ - \sin 30^\circ) = \frac{\mu_0 i}{4 \pi a (\sqrt{3}/2)} (\frac{\sqrt{3}-1}{2}) = \frac{\mu_0 i}{4 \pi a} \frac{\sqrt{3}-1}{\sqrt{3}} = \frac{\mu_0 i}{4 \pi a} (1 - \frac{1}{\sqrt{3}})$.Given the options,the correct expression is $C$.

Find the magnetic field at point $P$ in the given diagram. The wire carries a current $i$.

Similar Questions

$A$ loop carrying current $I$ has the shape of a regular polygon of $n$ sides. If $R$ is the distance from the centre to any vertex,then the magnitude of the magnetic induction vector $B$ at the centre of the loop is

Assertion: In electric circuits,wires carrying currents in opposite directions are often twisted together.
Reason: If the wires are not twisted together,the combination of the wires forms a current loop,and the magnetic field generated by the loop might affect adjacent circuits or components.

$A$ long thin hollow metallic cylinder of radius '$R$' carries a current '$i$'. The magnetic induction '$B$' at a distance '$r$' from the axis varies as shown in:

$A$ straight wire carrying a current of $12 \,A$ is bent into a semi-circular arc of radius $2 \,cm$ as shown in the figure. Then the magnetic field due to the straight segments at the centre of the arc is

Vedclass Test Series

Exam Paper Generator

Online Exam Module