A current carrying circular coil of radius R | vedclass

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

Question

(A) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_2 = \frac{\mu_0 I}{2R}$.The magnetic field a

Vedclass · Accepted Answer

$R \sqrt{3}$

Vedclass · Answer

(A) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_2 = \frac{\mu_0 I}{2R}$.The magnetic field at an axial point $P$ at a distance $x$ from the centre is given by $B_1 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Given $B_1 = \frac{B_2}{8}$,we substitute the expressions:$\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8} 	imes \frac{\mu_0 I}{2R}$.Simplifying the equation:$\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R}$.$8R^3 = (R^2 + x^2)^{3/2}$.Taking the power of $2/3$ on both sides:$(8R^3)^{2/3} = R^2 + x^2$.$4R^2 = R^2 + x^2$.$x^2 = 3R^2$.$x = R \sqrt{3}$.

$A$ 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

Similar Questions

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

Two concentric coils each of radius equal to $4 \pi \text{ cm}$ are placed at right angles to each other. If $10 \text{ A}$ and $24 \text{ A}$ are the currents flowing through the coils,respectively,the magnetic induction at the center of the coils will be

Two very long straight conductors (wires) are set parallel to each other. Each carries a current $I$ in the same direction and the separation between them is $2r$. The intensity of the magnetic field at point $P$ (as shown in the figure) ($\mu_0=$ permeability of free space) is

$A$ straight section $PQ$ of a circuit lies along the $X$-axis from $x = -\frac{a}{2}$ to $x = \frac{a}{2}$ and carries a steady current $i$. The magnetic field due to the section $PQ$ at a point $x = a$ on the $X$-axis will be:

Magnetic field at a distance $r$ from an infinitely long straight conductor carrying a steady current varies as

Vedclass Test Series

Exam Paper Generator

Online Exam Module