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

The magnetic field at a point $P$ situated at a perpendicular distance $R$ from a long straight wire carrying a current of $12 \ A$ is $3 \times 10^{-5} \ Wb/m^2$. The value of $R$ in $mm$ is $\left[\mu_0 = 4\pi \times 10^{-7} \ Wb/Am\right]$

$A$ circular coil carrying current has radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{27}$th of its value at the centre of the coil is:

Two concentric coils of $10$ turns each are placed in the same plane. Their radii are $20 \ cm$ and $40 \ cm$ and carry $0.2 \ A$ and $0.3 \ A$ current respectively in opposite directions. The magnetic induction (in $T$) at the centre is

In a hydrogen atom, an electron is revolving in an orbit of radius $0.53 \, \text{\AA}$ with $6.6 \times 10^{15} \, \text{rotations/second}$. The magnetic field produced at the centre of the orbit is ....... $Wb/m^2$.

The figure shows a straight wire of length $l$ carrying current $i$. The magnitude of the magnetic field produced by the current at point $P$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module