Discuss similarities and differences of Biot-Savart law with Coulomb’s law
Similarities and differences of Biot-Savart's law with Coulomb's law are as below. Similarities :
$(1)$ Both depend inversely on the square of distance.
$(2)$ Both are long range.
$(3)$ The principle of superposition applies to both fields.
Thus, static electric field $\mathrm{E}=\frac{k \mathrm{Q}}{r^{2}}$
$\therefore \mathrm{E} \propto \mathrm{Q}$
Similarly for Biot-Savart law,
$\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} d l \times \hat{r}}{r^{2}}$
$\therefore \mathrm{B} \propto \mathrm{I} d l$
Differences :
$(1)$ Magnetic field is produced due to vector component $\mathrm{I} d \vec{l}$. Whereas electric field is produced due to scalar component $d q$.
$(2)$ The electrostatic field is along the displacement vector joining the source and the field point, whereas, the magnetic field is perpendicular to current element $\mathrm{I} d \vec{l}$ and the plane containing the displacement vector $\vec{r}$.
$(3)$ Biot-Savart's law is depend on angle $(\sin \theta)$. The point on $\mathrm{I} d \vec{l}$ element be $\theta=0^{\circ}$ so that $\sin 0^{\circ}=0$ and no magnetic field on axial point whereas Coulomb's law does not depend on angle $\theta$.
Given below are two statements:
Statement $(I)$: When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.
Statement $(II)$: Concave lens always forms a virtual and erect image.
In the light of the above statements, choose the correct answer from the options given below:
Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field $B_1,$ at its centre when a current $I$ passes through it.The second wire is bent into a coil with three identical loops adjacent to each other and produces magnetic field $B_2$ at the centre of the loops when current $I/3$ passes through it. The ratio $B_1 : B_2$ is
A particle is moving with velocity $\overrightarrow{ v }=\hat{ i }+3 \hat{ j }$ and it produces an electric field at a point given by $\overrightarrow{ E }=2 \hat{ k }$. It will produce magnetic field at that point equal to (all quantities are in SI units)
.......$A$ should be the current in a circular coil of radius $5\,cm$ to annul ${B_H} = 5 \times {10^{ - 5}}\,T$