Two identical coils of radius R and number of | vedclass

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

Question

(D) The magnetic field at the center of a circular coil of radius $R$ with $N$ turns carrying current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.For th

Vedclass · Accepted Answer

$\frac{\mu_0 NI}{R}$

Vedclass · Answer

(D) The magnetic field at the center of a circular coil of radius $R$ with $N$ turns carrying current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.For the first coil with current $I_1 = I$, the magnetic field is $B_1 = \frac{\mu_0 NI}{2R}$.For the second coil with current $I_2 = I\sqrt{3}$, the magnetic field is $B_2 = \frac{\mu_0 N(I\sqrt{3})}{2R} = \sqrt{3} \frac{\mu_0 NI}{2R}$.Since the coils are placed perpendicular to each other, the magnetic fields $B_1$ and $B_2$ are also perpendicular to each other $(	heta = 90^{\circ})$.The resultant magnetic field $B_{	ext{net}}$ is given by:$B_{	ext{net}} = \sqrt{B_1^2 + B_2^2}$Substituting the values:$B_{	ext{net}} = \sqrt{\left(\frac{\mu_0 NI}{2R}ight)^2 + \left(\sqrt{3} \frac{\mu_0 NI}{2R}ight)^2}$$B_{	ext{net}} = \frac{\mu_0 NI}{2R} \sqrt{1^2 + (\sqrt{3})^2}$$B_{	ext{net}} = \frac{\mu_0 NI}{2R} \sqrt{1 + 3} = \frac{\mu_0 NI}{2R} \sqrt{4} = 2 \left(\frac{\mu_0 NI}{2R}ight) = \frac{\mu_0 NI}{R}$.

Two identical coils of radius $R$ and number of turns $N$ are placed perpendicular to each other such that they have a common center. The currents through them are $I$ and $I\sqrt{3}$. The resultant intensity of the magnetic field at the center of the coils will be:

Similar Questions

$A$ particle is moving with velocity $\overrightarrow{v} = \hat{i} + 3\hat{j}$ and it produces an electric field at a point given by $\overrightarrow{E} = 2\hat{k}$. It will produce a magnetic field at that point equal to (all quantities are in $SI$ units):

$A$ current carrying circular coil of radius $R$ produces magnetic fields $B_1$ and $B_2$ at an axial point $P$ at a distance $x$ from its centre and at a point $Q$ placed at its centre respectively. If $B_1 = \frac{B_2}{8}$,the value of $x$ is

$A$ current of $1.5 \, A$ is flowing through an equilateral triangle of side $9 \, cm$. The magnetic field at the centroid of the triangle is (Assume that the current is flowing in the clockwise direction.)

$A$ current of $1\,A$ is passed through a hexagonal conducting wire of side $1\,m$. The magnetic induction at its centre $O$ in $Wb/m^2$ will be

The magnetic induction due to an infinitely long straight wire carrying a current $i$ at a distance $r$ from the wire is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module