Biot-savart's law - class-XI

Electric current and magnetism are very closely related phenomenon and both can be generated using the other.

Electromagnetism is study of various phenomenon related to electricity and magnetism. 

Electric current and magnetism are very closely related phenomenon and both can be generated using the other.

Direction of magnetic field lines around a current carrying conductor can be found by RIght Hand Thumb Rule.

A charged particle produces an electric field around it independent of its motion. But when it is at rest, it doesn't produce a magnetic field. 

According to the right-hand thumb rule, the magnetic field lines around a long straight current-carrying conductor are concentric circles centered on the wire.

Coulombic force between two charges is given by

F=1/4π€ (q'.q″/r²)

From here, the SI UNIT of permittivity comes out is

= coulomb²/Newton-meter²

C²/N-m²

According to Biot-Savart's law, the magnetic field
$\displaystyle \overset{\rightarrow}{B} = \frac{\mu _o}{4 \pi} . \frac{q (\overset{\rightarrow}{v} \times \overset{\rightarrow}{r} ) }{r^3}$
The direction of $\overset{\rightarrow}{B}$ will be along $\overset{\rightarrow}{v} \times \overset{\rightarrow}{r}$ i.e. perpendicular to the plane containing $\overset{\rightarrow}{v}$ and $\overset{\rightarrow}{r}$.

Magnetic field produced by charges moving with velocity $\bar v$, at a distance r is $ \bar B$ = $\left ( \dfrac{\mu _0}{4\pi } \right )$.q$\dfrac{\bar v \times \hat r}{r^2}$
Therefore $\bar B \perp \bar v$

The particle moves in a circular path in a magnetic field. The radius of the path is r = v0 / (alpha * B0). Since it starts at the origin and moves along the x-axis, it reaches (0, y, 0) after completing a semicircle, so y = 2r = 2 * v0 / (alpha * B0).

As per Biot Savart Law, Magnetic field at a point is inversely proportional to square of distance from the current carrying conductor. Therefore magnitude of magnetic field is same at both points P and Q, irrespective of their position from the conductor.

The Magnitude of magnetic field produced by a straight current carrying wire at a given point is
A) directly proportional to the current passing in the wire and
B) inversely proportional to the distance of that point from the wire.

The magnetic field produced by a current-carrying wire at a given point depends directly on the current passing through it. Therefore, if there is a circular coil having n turns, the field produced is n times as large as that produced by a single turn. This is because the current in each circular turn has the same direction, and the field due to each turn then just adds up.

The Magnitude of magnetic field produced by a straight current carrying wire at a given point is 
A) directly proportional to the current passing in the wire and 
B) inversely proportional to the distance of that point from the wire.

If unit current is flowing through the conductor, then Biot-Savart's law is represented as
    $\vec { dB } =\dfrac { { \mu  } _{ 0 } }{ 4\pi  } \dfrac { \vec { dl } \times \vec { r }  }{ { r }^{ 3 } } $

Let the current $i$ is uniformly distributed throughout the wire.

Inside a charging capacitor, the magnetic field at distance r from the axis is proportional to r. Therefore, the ratio of the field at R/2 to the field at R is (R/2) / R = 1/2.

$B= \dfrac{\mu _0}{4\pi}i\dfrac{\vec{dl}\times  \vec{r}}{r^3}=-\dfrac{\mu _0}{4\pi}i\dfrac{\vec{r}\times  \vec{dl}}{r^3}$
since $\vec{dl}\times  \vec{r} = -(\vec{r}\times  \vec{dl})$
Options (A) and (C) are correct

The radius of the path is r = mv / (qB). For the same velocity and distance, the particle with the smallest radius (highest q/m ratio) will deviate the most. Li++ has a higher charge-to-mass ratio than the others.

Biot-Savart's law states that the magnetic field dB is proportional to idlsin(theta)/r^2. The constant of proportionality is mu0/4pi.

Upon completing a full circle net displacement is 0.
Work done by the magnetic field is 0 because the net displacement caused by the magnetic field is 0.

$B=\dfrac{\mu _o}{4\pi}\dfrac{i\overrightarrow{dl}\times\overrightarrow{r}}{r^3}$

So it depends on all three : current , distance and length
but not on nature of material

Bio-savert law - apply directly the statement of law

because angle between line and distance becomes zero.
$\therefore\overrightarrow{l}\times\widehat{r}$  becomes zero.

Answer is D.

The magnitude and direction of F depend on the velocity of the particle and on the magnitude and direction of the magnetic field B.
When a charged particle moves parallel to the magnetic field vector, the magnetic force acting on the particle is zero.
When the particles velocity vector makes any angle with the magneticfield, the magnetic force acts in a direction perpendicular to both v and B; that is, F is perpendicular to the plane formed by v and B.
The magnitude of the magnetic force is
$F=qvBsin\theta $
where $\theta $ is the smaller angle between v and B. From this expression, we see that F is zero when v is parallel or antiparallel to B or 180) and maximum when v is perpendicular to B.
In this case, as the proton moves parallel to the magnetic field, the force is zero.

Magnetic field due to large wire is 

The physical constant $μ _0$ commonly called the vacuum permeability, permeability of free space, ... In the reference medium of classical vacuum, μ0 has an exact defined value: .... The value of $μ _0$ was chosen such that the rmks unit of current is equal in size to the ampere in the emu system: μ0 is defined to be $4π × 10^{−7} H/m.$

The Biot Savart law is used for computing the resultant magnetic field B at position r generated by a steady current I.  According to it $B=\frac {\mu _0 I } {4\pi} \int \frac {I\vec{dl} \times \vec{r}} { r^3}$.
So magnetic field at a distance r and lying on the plane perpendicular ($\pi /2$)to the element of conductor is maximum .

The magnetic field due to a long straight conductor is given by the Biot-Savart law, which leads to the formula B = mu0i / (2pi*r).