As shown in the figure, a current of $2\,A$ flowing 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 earth's magnetic field.)
A$4 \sqrt{3} \times 10^{-4} \,T$
B$4 \sqrt{3} \times 10^{-5} \,T$
C$\sqrt{3} \times 10^{-4}\, T$
D$3 \sqrt{3} \times 10^{-5}\,T$
JEE MAIN 2023, Medium
Download our app for free and get started
D$3 \sqrt{3} \times 10^{-5}\,T$
d $d \tan 60^{\circ}=2 \sqrt{3}$
$d =2\,cm$
$B =3 \times \frac{\mu_0 i }{2 \pi d } \sin 60^{\circ}$
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
For a solenoid keeping the turn density constant its length is halved and its cross section radius is doubled then the inductance of the solenoid increased by.....$\%$
A long solenoid has a radius $a$ and number of turns per unit length is $n$. If it carries a current $i$, then the magnetic field on its axis is directly proportional to
The flux density obtained at the centre of a circular coil of radius $R$ which carries a current $i$, is $B_0$. At a distance $‘pR’$ from the centre on the axis, the flux density will be
$PQ$ and $RS$ are long parallel conductors separated by certain distance. $M$ is the mid-point between them (see the figure). The net magnetic field at $M$ is $B$ . Now, the current $2\,A$ is switched off. The field at $M$ now becomes
In a galvanometer, the deflection becomes one half when the galvanometer is shunted by a $20\,\,\Omega$ resistor. The galvanometer resistance is ............... $\Omega$