A long thin-walled pipe of radius R carries a | vedclass

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

Question

(A) Consider a small element of the pipe subtending an angle $d	heta$ at the center. The current in this element is $dI = \frac{I}{2\pi} d	heta$.The

Vedclass · Accepted Answer

$\frac{{\mu _0}I\sqrt 2 }{4{\pi ^2}R}$

Vedclass · Answer

(A) Consider a small element of the pipe subtending an angle $d	heta$ at the center. The current in this element is $dI = \frac{I}{2\pi} d	heta$.The magnetic field $dB$ at the center due to this long current-carrying strip (acting as an infinite wire) is given by $dB = \frac{\mu_0 (dI)}{2\pi R} = \frac{\mu_0 I}{4\pi^2 R} d	heta$.The direction of $dB$ is perpendicular to the radius vector of the element. For the quarter portion,the angle $	heta$ ranges from $0$ to $\pi/2$. Resolving the field into components:$dB_x = dB \sin	heta$ and $dB_y = -dB \cos	heta$.Integrating these components from $0$ to $\pi/2$:$B_x = \int_0^{\pi/2} \frac{\mu_0 I}{4\pi^2 R} \sin	heta d	heta = \frac{\mu_0 I}{4\pi^2 R} [-\cos	heta]_0^{\pi/2} = \frac{\mu_0 I}{4\pi^2 R}$.$B_y = \int_0^{\pi/2} \frac{\mu_0 I}{4\pi^2 R} \cos	heta d	heta = \frac{\mu_0 I}{4\pi^2 R} [\sin	heta]_0^{\pi/2} = \frac{\mu_0 I}{4\pi^2 R}$.The magnitude of the resultant magnetic field is $B = \sqrt{B_x^2 + B_y^2} = \sqrt{(\frac{\mu_0 I}{4\pi^2 R})^2 + (\frac{\mu_0 I}{4\pi^2 R})^2} = \frac{\mu_0 I \sqrt{2}}{4\pi^2 R}$.

$A$ long thin-walled pipe of radius $R$ carries a current $I$ along its length. The current density is uniform over the circumference of the pipe. The magnetic field at the center of the pipe due to the quarter portion of the pipe shown is:

Similar Questions

As the temperature of the hot junction increases, the thermo $e.m.f.$

The figure shows a circular coil carrying current $i$ kept very close but not touching at a point $A$ on a straight conductor carrying the same current $i$. The magnitude of magnetic induction at the centre $O$ of the circular coil will be:

The dimensions of $\left(\frac{B^{2}}{\mu_{0}}\right)$ will be. (where $\mu_{0}$ is the permeability of free space and $B$ is the magnetic field)

$A$ rectangular coil carrying current is placed in a non-uniform magnetic field. On that coil, the total:

Vedclass Test Series

Exam Paper Generator

Online Exam Module