As shown in the figure,a current of 2,A flows | vedclass

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

Question

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

Vedclass · Accepted Answer

$3 \sqrt{3} 	imes 10^{-5} \,T$

Vedclass · Answer

(D) The magnetic field due to a finite straight wire at a perpendicular distance $d$ is given by $B = \frac{\mu_0 i}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2)$.For an equilateral triangle,the distance $d$ from the centroid to any side is $d = \frac{a}{2 \sqrt{3}}$,where $a = 4 \sqrt{3} \,cm = 4 \sqrt{3} 	imes 10^{-2} \,m$.Thus,$d = \frac{4 \sqrt{3} 	imes 10^{-2}}{2 \sqrt{3}} = 2 	imes 10^{-2} \,m$.The angles at the centroid subtended by the ends of each side are $	heta_1 = 	heta_2 = 60^{\circ}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi d} (\sin 60^{\circ} + \sin 60^{\circ}) = \frac{\mu_0 i}{4 \pi d} (2 \sin 60^{\circ}) = \frac{\mu_0 i}{2 \pi d} \sin 60^{\circ}$.The total magnetic field at the centroid due to three sides is $B = 3 	imes B_1 = 3 	imes \frac{\mu_0 i}{2 \pi d} \sin 60^{\circ}$.Substituting the values: $B = 3 	imes \frac{2 	imes 10^{-7} 	imes 2}{2 	imes 10^{-2}} 	imes \frac{\sqrt{3}}{2} = 3 	imes 2 	imes 10^{-5} 	imes \frac{\sqrt{3}}{2} = 3 \sqrt{3} 	imes 10^{-5} \,T$.

As shown in the figure,a current of $2\,A$ flows in an equilateral triangle of side $4 \sqrt{3}\,cm$. The magnetic field at the centroid $O$ of the triangle is:
(Neglect the effect of the earth's magnetic field.)

Similar Questions

$A$ straight wire of diameter $0.5\, mm$ carrying a current of $1\, A$ is replaced by another wire of $1\, mm$ diameter carrying the same current. The strength of the magnetic field at a point far away is

$A$ current of $0.1\, A$ circulates around a coil of $100$ turns and having a radius equal to $5\, cm$. The magnetic field set up at the centre of the coil is $(\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m)$

$A$ circular coil of wire consisting of $n$ turns,each of radius $8 \ cm$,carries a current of $0.4 \ A$. The magnitude of the magnetic field at the centre of the coil is $3.14 \times 10^{-4} \ T$. The value of $n$ is (Take $\mu_0 = 12.56 \times 10^{-7} \ T \cdot m/A$)

An electron of mass $m$ is revolving around the nucleus in a circular orbit of radius $r$ and has angular momentum $L$. The magnetic field produced by the electron at the centre of the orbit is ($e = \text{electric charge}$, $\mu_{0} = \text{permeability of free space}$)

An electron is revolving around a proton, producing a magnetic field of $16 \, Wb/m^2$ in a circular orbit of radius $1 \, Å$. Its angular velocity will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

As shown in the figure,a current of $2\,A$ flows in an equilateral triangle of side $4 \sqrt{3}\,cm$. The magnetic field at the centroid $O$ of the triangle is:(Neglect the effect of the earth's magnetic field.)