For the adjoining figure,the magnetic field a | vedclass

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

Question

(D) The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$.For the wire carrying $5 	ext{ A}$ current at a distance

Vedclass · Accepted Answer

$\frac{\mu_0}{2 \pi} \otimes$

Vedclass · Answer

(D) The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$.For the wire carrying $5 	ext{ A}$ current at a distance of $2.5 	ext{ m}$ from $P$,the magnetic field is:$B_1 = \frac{\mu_0 	imes 5}{2 \pi 	imes 2.5} = \frac{2 \mu_0}{2 \pi} \otimes$ (directed into the page).For the wire carrying $2.5 	ext{ A}$ current,the distance from $P$ is $5 	ext{ m} - 2.5 	ext{ m} = 2.5 	ext{ m}$. The magnetic field is:$B_2 = \frac{\mu_0 	imes 2.5}{2 \pi 	imes 2.5} = \frac{\mu_0}{2 \pi} \odot$ (directed out of the page).The net magnetic field at $P$ is:$B_{net} = B_1 - B_2 = \frac{2 \mu_0}{2 \pi} - \frac{\mu_0}{2 \pi} = \frac{\mu_0}{2 \pi} \otimes$.

For the adjoining figure,the magnetic field at point $P$ will be:

Similar Questions

$A$ long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $B$. It is then bent into a circular loop of $n$ turns. The magnetic field at the centre of the coil for the same current will be

The magnetic field at the point of intersection of diagonals of a square wire loop of side $L$ carrying a current $I$ is

$A$ circular coil of wire consisting of $100$ turns,each of radius $8.0 \, cm$,carries a current of $0.40 \, A$. What is the magnitude of the magnetic field $B$ at the centre of the coil?

Vedclass Test Series

Exam Paper Generator

Online Exam Module