The magnetic field at the origin due to the c | vedclass

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

Question

(C) The wire consists of two semi-infinite segments. Segment $OD$ lies along the $x$-axis. The origin $O$ lies on the axis of this wire segment,so the

Vedclass · Accepted Answer

$\frac{{\mu _0 I}}{{8\pi a}}\left( { - \hat i + \hat k} \right)$

Vedclass · Answer

(C) The wire consists of two semi-infinite segments. Segment $OD$ lies along the $x$-axis. The origin $O$ lies on the axis of this wire segment,so the magnetic field due to segment $OD$ at the origin is $B_{OD} = 0$. Segment $AB$ is parallel to the $y$-axis and passes through the point $(a, 0, a)$. The distance of the origin from this wire is $r = \sqrt{a^2 + a^2} = a\sqrt{2}$. The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is $B = \frac{\mu_0 I}{4\pi r}$. Using the right-hand rule,the direction of the magnetic field at the origin due to segment $AB$ is perpendicular to the plane containing the wire and the origin. The vector from the wire to the origin is along $(-\hat{i} - \hat{k})$. The current is along $\hat{j}$. The direction is $\hat{j} 	imes (\hat{i} + \hat{k}) = -\hat{k} + \hat{i}$. Normalizing the direction vector,the field is $B = \frac{\mu_0 I}{4\pi (a\sqrt{2})} \frac{(-\hat{i} + \hat{k})}{\sqrt{2}} = \frac{\mu_0 I}{8\pi a} (-\hat{i} + \hat{k})$.

The magnetic field at the origin due to the current $I$ flowing in the wire as shown in the figure is:

Similar Questions

Two long parallel wires are at a distance $2d$ apart. They carry steady equal currents flowing out of the plane of the paper,as shown. The variation of the magnetic field $B$ along the line $XX'$ is given by

$A$ coil having $N$ turns carries a current $I$ as shown in the figure. The magnetic field intensity at point $P$ is

Two concentric circular coils of ten turns each are situated in the same plane. Their radii are $20 \ cm$ and $40 \ cm$ and they carry respectively $0.2 \ A$ and $0.3 \ A$ current in opposite directions. The magnetic field in $Wb/m^2$ at the centre is:

An arc of a circle of radius $R$ subtends an angle $\frac{\pi}{2}$ at the centre. It carries a current $I$. The magnetic field at the centre will be ($\mu_0 =$ permeability of free space).

$A$ coil having $100$ turns is wound tightly in the form of a spiral with inner and outer radii $1 \text{ cm}$ and $2 \text{ cm}$,respectively. When a current $1 \text{ A}$ passes through the coil,the magnetic field at the centre of the coil is

Vedclass Test Series

Exam Paper Generator

Online Exam Module