Shown in the figure is a conductor carrying a | vedclass

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

Question

(A) The magnetic field due to an arc of radius $R$ subtending an angle $	heta$ at the center is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$.In the

Vedclass · Accepted Answer

$\frac {5\mu _0I	heta }{24\,\pi r}$

Vedclass · Answer

(A) The magnetic field due to an arc of radius $R$ subtending an angle $	heta$ at the center is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$.In the given figure,there are three arcs with radii $r$,$2r$,and $3r$ respectively,all subtending the same angle $	heta$ at point $O$.Using the right-hand rule,the magnetic field due to the arc of radius $r$ is directed into the page,the arc of radius $2r$ is directed out of the page,and the arc of radius $3r$ is directed into the page.Let $B_1, B_2, B_3$ be the magnetic fields due to the arcs of radii $r, 2r, 3r$ respectively.$B_1 = \frac{\mu_0 I 	heta}{4 \pi r}$ (inward)$B_2 = \frac{\mu_0 I 	heta}{4 \pi (2r)} = \frac{\mu_0 I 	heta}{8 \pi r}$ (outward)$B_3 = \frac{\mu_0 I 	heta}{4 \pi (3r)} = \frac{\mu_0 I 	heta}{12 \pi r}$ (inward)The net magnetic field $B$ at $O$ is $B = B_1 - B_2 + B_3$.$B = \frac{\mu_0 I 	heta}{4 \pi} \left[ \frac{1}{r} - \frac{1}{2r} + \frac{1}{3r} ight]$$B = \frac{\mu_0 I 	heta}{4 \pi r} \left[ 1 - \frac{1}{2} + \frac{1}{3} ight]$$B = \frac{\mu_0 I 	heta}{4 \pi r} \left[ \frac{6 - 3 + 2}{6} ight] = \frac{\mu_0 I 	heta}{4 \pi r} \left[ \frac{5}{6} ight] = \frac{5 \mu_0 I 	heta}{24 \pi r}$.

Shown in the figure is a conductor carrying a current $I$. The magnetic field intensity at the point $O$ (common centre of all the three arcs) is

Similar Questions

Identical currents flow in two perpendicular wires,as shown in the figure. The wires are very close but do not touch. The magnetic field can be zero:

The Earth's magnetic field at a given point is $0.5 \times 10^{-5} \, Wb/m^2$. This field is to be annulled by the magnetic induction at the center of a circular conducting loop of radius $5.0 \, cm$. The current required to be flown in the loop is nearly......$A$

$A$ conducting wire is in the shape of a regular hexagon which is inscribed inside an imaginary circle of radius $R$. If a current $I$ flows through the wire, the magnitude of the magnetic field at the centre of the circle is

The figure shows a current-carrying square loop $\text{ABCD}$ of edge length '$a$' lying in a plane. If the resistance of the $\text{ABC}$ part is $r$ and that of the $\text{ADC}$ part is $2r$,then the magnitude of the resultant magnetic field at the centre of the square loop is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module