The magnetic field at the origin due to a cur | vedclass

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

Question

(B) According to the Biot-Savart Law,the magnetic field $d\vec{B}$ at a point $P$ due to a current element $i \, d\vec{l}$ located at a position vecto

Vedclass · Accepted Answer

$(ii), (iii)$

Vedclass · Answer

(B) According to the Biot-Savart Law,the magnetic field $d\vec{B}$ at a point $P$ due to a current element $i \, d\vec{l}$ located at a position vector $\vec{r}'$ relative to the source is given by $d\vec{B} = \frac{\mu_0}{4\pi} \frac{i \, d\vec{l} 	imes \vec{r}_{rel}}{r_{rel}^3}$,where $\vec{r}_{rel}$ is the vector from the source to the point of observation.Here,the current element is at position $\vec{r}$ and we need the field at the origin $(0,0,0)$. Thus,the vector from the source to the origin is $\vec{r}_{rel} = \vec{0} - \vec{r} = -\vec{r}$.Substituting this into the formula:$d\vec{B} = \frac{\mu_0 i}{4\pi} \frac{d\vec{l} 	imes (-\vec{r})}{r^3} = -\frac{\mu_0 i}{4\pi} \frac{d\vec{l} 	imes \vec{r}}{r^3}$. This matches expression $(ii)$.Using the property of cross products,$d\vec{l} 	imes \vec{r} = -(\vec{r} 	imes d\vec{l})$.Substituting this into the expression for $d\vec{B}$:$d\vec{B} = -\frac{\mu_0 i}{4\pi} \frac{-(\vec{r} 	imes d\vec{l})}{r^3} = \frac{\mu_0 i}{4\pi} \frac{\vec{r} 	imes d\vec{l}}{r^3}$. This matches expression $(iii)$.Therefore,expressions $(ii)$ and $(iii)$ are correct.

Similar Questions

Two concentric coils each of radius equal to $2\pi \text{ cm}$ are placed at right angles to each other. If $3 \text{ A}$ and $4 \text{ A}$ are the currents flowing through the two coils respectively,the magnetic induction (in $\text{Wb m}^{-2}$) at the centre of the coils will be:

The current is flowing in the south direction along a power line. The direction of the magnetic field above the power line (neglecting the Earth's field) is

The wire loop $PQRSP$ formed by joining two semicircular wires of radii $R_1$ and $R_2$ carries a current $I$ as shown. The magnitude of the magnetic field at the centre '$O$' is

$A$ closely wound circular coil of radius $5\,cm$ produces a magnetic field of $37.68 \times 10^{-4}\,T$ at its center. The current through the coil is $......A$. [Given,number of turns in the coil is $100$ and $\pi=3.14$]

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module