The magnetic field doverrightarrow{B} due to | vedclass

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

Question

(D) According to the Biot-Savart Law,the magnetic field $dB$ due to a current element $idl$ at a distance $r$ is given by the magnitude: $dB = \frac{\

Vedclass · Accepted Answer

$d\overrightarrow{B} = \frac{\mu_0}{4\pi} i \left( \frac{d\overrightarrow{l} 	imes \overrightarrow{r}}{r^3} ight)$

Vedclass · Answer

(D) According to the Biot-Savart Law,the magnetic field $dB$ due to a current element $idl$ at a distance $r$ is given by the magnitude: $dB = \frac{\mu_0}{4\pi} \frac{idl \sin 	heta}{r^2}$.In vector form,this is expressed as $d\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{i(d\overrightarrow{l} 	imes \widehat{r})}{r^2}$.Since the unit vector $\widehat{r} = \frac{\overrightarrow{r}}{r}$,we substitute this into the equation:$d\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{i(d\overrightarrow{l} 	imes \overrightarrow{r})}{r^2 \cdot r} = \frac{\mu_0}{4\pi} \frac{i(d\overrightarrow{l} 	imes \overrightarrow{r})}{r^3}$.Thus,option $D$ is correct.

The magnetic field $d\overrightarrow{B}$ due to a small current element $d\overrightarrow{l}$ at a distance $\overrightarrow{r}$ from an element carrying current $i$ is given by:

Similar Questions

Write the formula for the magnetic field due to a circular current-carrying loop having $N$ turns and $R$ radius at a point on the axis of the loop at a distance $x$ from the center.

The magnetic field at a perpendicular distance '$r$' from a long straight wire carrying current '$I$' is '$B$'. The magnetic field at a perpendicular distance '$2r$' from the same wire is:

Which rule can be used to determine the direction of the magnetic field due to a current-carrying loop?

The magnetic field at the center of a circular coil of radius $0.1\, m$ having $1000$ turns and carrying a current of $0.1\, A$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module