Biot-Savart law indicates that the moving ele | vedclass

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

Question

(A) According to the Biot-Savart law for a moving point charge $q$ with velocity $\vec{v}$,the magnetic field $\vec{B}$ at a position vector $\vec{r}$

Vedclass · Accepted Answer

$\vec{B} \perp \vec{v}$

Vedclass · Answer

(A) According to the Biot-Savart law for a moving point charge $q$ with velocity $\vec{v}$,the magnetic field $\vec{B}$ at a position vector $\vec{r}$ is given by:$\vec{B} = \frac{\mu_0}{4\pi} \frac{q(\vec{v} 	imes \vec{r})}{r^3}$Since the magnetic field $\vec{B}$ is proportional to the cross product of the velocity vector $\vec{v}$ and the position vector $\vec{r}$,the resulting vector $\vec{B}$ must be perpendicular to both $\vec{v}$ and $\vec{r}$.Therefore,$\vec{B} \perp \vec{v}$.

Biot-Savart law indicates that the moving electrons (velocity $\vec{v}$) produce a magnetic field $\vec{B}$ such that

Similar Questions

The Earth's magnetic induction at a certain point is $7 \times 10^{-5} \ Wb/m^2$. This is to be annulled by the magnetic induction at the center of a circular conducting loop of radius $5 \ cm$. The required current in the loop is......$A$

The magnetic field at the centre $C$ of the circular arc is:

$A$ current $I$ is flowing in a conductor $PQRST$ as shown in the figure. The radius of the curved path $QRS$ is $r$ and the length of the straight portions $PQ$ and $ST$ is very large. The magnetic field at the centre $O$ of the curved part is $(\mu_0 = \text{permeability of free space})$

The magnetic induction at the centre $O$ is:

$A$ long thin hollow metallic cylinder of radius '$R$' carries a current '$i$'. The magnetic induction '$B$' at a distance '$r$' from the axis varies as shown in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module