A long straight wire carrying a constant curr | vedclass

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

Question

(A) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $i$ is given by $B = \frac{\mu_0 i}{2 \pi r}$.The area $A$ und

Vedclass · Accepted Answer

$Al$

Vedclass · Answer

(A) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $i$ is given by $B = \frac{\mu_0 i}{2 \pi r}$.The area $A$ under the $B-r$ curve between $r_1$ and $r_2$ is defined as $A = \int_{r_1}^{r_2} B \, dr = \int_{r_1}^{r_2} \frac{\mu_0 i}{2 \pi r} \, dr = \frac{\mu_0 i}{2 \pi} \ln\left(\frac{r_2}{r_1}\right)$.The magnetic flux $\phi$ through the rectangular loop of length $l$ is given by $\phi = \int B \, dA_{loop}$.Considering a strip of width $dr$ at a distance $r$ from the wire,the area of the strip is $dA_{loop} = l \, dr$.Thus,the flux is $\phi = \int_{r_1}^{r_2} B \, l \, dr = l \int_{r_1}^{r_2} B \, dr$.Since $A = \int_{r_1}^{r_2} B \, dr$,we substitute this into the flux equation:$\phi = l \cdot A = Al$.

Similar Questions

If the magnetic flux is expressed in $Weber$,then magnetic induction can be expressed in

$A$ square of side $x \, m$ lies in the $x-y$ plane in a region where the magnetic field is given by $\vec B = B_0 (3\hat i + 4\hat j + 5\hat k ) \, T$,where $B_0$ is a constant. The magnitude of the magnetic flux passing through the square is:

$A$ circular disc of radius $0.2 \; m$ is placed in a uniform magnetic field of induction $\frac{1}{\pi} \; Wb/m^2$ in such a way that its axis makes an angle of $60^{\circ}$ with $\vec{B}$. The magnetic flux linked with the disc is..... $Wb$.

Imagine rolling a sheet of paper into a cylinder and placing a bar magnet near its end as shown in the figure. What can you say about the sign of $\vec B \cdot d\vec A$ for every area element $d\vec A$ on the surface of the cylinder?

$A$ circular disc of radius $20 \text{ cm}$ is placed in a uniform magnetic field of induction $\frac{7}{22} \text{ Wb/m}^2$ in such a way that its axis makes an angle of $60^{\circ}$ with $\vec{B}$. The magnetic flux linked with the disc is $(\cos 60^{\circ} = 0.5)$ (in $\text{ Wb}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module