Magnetic Effect Of Current

Magnetic Effect of Current

Oersted's Experiment

A magnetic field is produced in the surrounding of any current carrying conductor.

The direction of this magnetic field can be obtained by Ampere's swimming ruel.

SI unit of magnetic field is Wm-2 or T (tesla).

The strength of magnetic field is called one tesla, if a charge of one coulomb, when moving with a velocity of 1 ms-1 along a direction perpendicular to the direction of the magnetic field experiences a force of one newton.

1 tesla (T) = 1 weber metre-2 (Wbm-2)

= 1 newton ampere-1 metre-1 (NA-1 m-1)

CGS units of magnetic field are called gaues or oersted.

1 gauss = 10-4 tesla.

Maxwell's Cork Screw Rule

if a right handed cork screw is imagined to be rotated in such a direction that tip of screw points in the direction of the current then direction of rotation of corkscrew gives the direction of magnetic line of force.

The conventional sign for a magnetic field coming out of the plane and normal to it is denoted by $\odot .$

The magnetic field perpendicular to the plane in the downward direction is denoted by $\otimes .$

Ampere's Swimming Rule

If a man is swimming along the wire in the direction of current with his face turned towards the needle, so that the current enters through his feet, then north pole of the magnetic needle will be deflected towards his left hand.

Magnetic Field

The space in the surrounding of a magnet of any current carrying conductor in which its magnetic influence can be experienced.

Biot Savart's Law

The magnetic field produced by a current carrying element of length dl, carrying current I at a point separated by a distance r is given by

$d\text{B}=\frac{{\mu }_0}{4\pi }\frac{I\left(d\text{l}\times \text{r}\right)}{r^3}$

or $dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$

where, $\theta$is the angle between the direction of the current and ${\mu }_0$ is absolute permeability of the free space.

SI unit of magnetic field is Wm-2 or (tesla) and CGS unit of magnetic field is gauss or oersted 1 gauss = 10-4 tesla.

The direction of magnetic field $d\text{B}$ is that of $Id\text{l}\times \text{r}$.

Magnetic Field Due to Straight Current Carrying Conductor

$B=\frac{{\mu }_0}{4\pi }\frac{I}{r}\left(\sin {\varphi }_1+\sin {\varphi }_2\right)$

Where, ${\varphi }_{1}and{\varphi }_2$ are angels, which the lines joining the two ends of the conductor to the observation point make with the perpendicular form the observation point to the conductor.

For infinite length conductor and observation point is near the centre of the conductors,

$B=\frac{{\mu }_0}{4\pi }\frac{2I}{r}$

For infinite length conductor and observation point is near one end of the conducotors,

$B=\frac{{\mu }_0}{4\pi }\frac{I}{r}$

Magnetic Field Lines

• They are used to represent magnetic field in a region.

• They are closed continuous curves.

• Tangent drawn at any point gives the direction of magnetic field.

• They cannot interact.

• Outside a magnet, they are directed from north to south pole and inside a magnet they are directed from south to north.

The magnetic field lines due to a straight current carrying conductor are concentric circles having centre at conductor and in a plane perpendicular to the conductor.

The direction of magnetic field lined can be obtained by Right Hand Thumb Rule.

Right Hand Thumb Rule

If we hold a current carrying conductor in the grip of the right hand in such a way that thumb points in the direction of current, then curling of fingers represents the direction of magnetic field lines.

Magnetic Field on the Axis of a Current Carrying Circular Coil

Magnetic field at axis at a distance x from centre O.

$B=\frac{{\mu }_{0}nI{r}^2}{2{\left(r^2+x^2\right)}^{3/2}}$

Where, r = radius of the coil, n = number of turns in the coil and

I = Current,

At centre of the coil, $B=\frac{{\mu }_{0}nI}{2r}$

If we look at one face of the coil and the direction of current flowing through the coil is clockwise, then that face has south polarity and if direction of current is anti-clockwise, then that face has north polarity.

Magnetic Dipole

Every current carrying loop is a magnetic dipole. It has two poles south (S) and north (N).

This is similar to a bar magnet.

Each magnetic dipole has some magnetic moment (M). The magnitude of M is,

$\left|M\right|=NiA$

Where, N = number of turns in the loop,

i = current in the loop and

A = area of cross-section of the loop.

The current carrying loop behaves as a small magnetic dipole placed along the axis one face of the loop behaves as north pole while the other face of loop behaves as South pole.

Ampere's Circular Law

The line integral of magnetic field induction B around any closed path in vacuum is equal to ${\mu }_0$ times the total current threading the closed path, i.e.

${\oint \text{B}.d\text{I}=\mu }_{0}I$

Where B is the magnetic field, d1 is small element, ${\mu }_0$ is the absolute permeability of free space and I is the current.

Ampere's circuital law holds good for a closed path of any size and shape around a current carrying conductor because the relation is independent of distance form conductor.

Solenoid

A solenoid is a closely wound helix or insulated copper wire.

Magnetic field at a point well inside a long solenoid is given by

$B={\mu }_{0}nI$

Where, n = number of turns per unit length and

I = current flowing through the solenoid.

Magnetic field at a point on one end of a long solenoid is given by

$B=\frac{{\mu }_{0}nI}{2}$

Toroid

A toroidal solenoid is an anchor ring around which is large number of turns of a copper wire are wrapped.

A toroid is an endless solenoid in the form of a ring.

Magentic field inside the turns of toroid is given by

$B={\mu }_{0}nI$

Magnetic field inside a toroid is constant and is always tangential to the circular closed path.

Magnetic field at any point inside the empty space surrounded by the toroid and outside the toroid, is zero, because net current enclosed by these space is zero.

Magnetic Field Due to a Current Carrying Long Circular Cylinder

Outside the cylinder (r > R)

$B=\frac{{\mu }_0}{2\pi }\frac{I}{r}$

Inside the cylinder when it is made of a thin metal sheet,

B = 0

Inside the cylinder when current is uniformly distributed throughout the cross-section of the cylinder (r > R)

$B=\frac{{\mu }_0{\mu }_r}{2\pi }\frac{Ir}{R^2}$

Where, ${\mu }_0$ and ${\mu }_r$ are permeabilities of free space and material of the cylinder, I is current flowing through the cylinder and r is radius of the cylinder.]

Force Acting on a Charge Particle Moving in a Uniform Magnetic Field

F = q (v × B)

or $F=\left|\text{F}\right|=Bqv\sin \theta$

where, B = magnetic field intensity

q = charge on particle

v = speed of the particle and

$\theta$ = angle between magnetic field and direction of motion.

This force is perpendicular to B as well as v.

Its direction can be obtained from Fleming's left hand rule.

Magnetic force acing on a current carrying conductor in a uniform magnetic field is given by

F=I (L x B)

or $F=BI\text{l}\sin \theta$

Fleming's Left Hand Rule

If we stretch the thumb, the forefinger and the central finger of left hand in such a way that all three are perpendicular to each other, then if forefinger represents the direction of magnetic field, central finger represents the direction of current flowing through the conductor, then thumb will represent the direction for magnetic force.

Lorentz Force

The total force experienced by a charge moving inside the electric and magnetic fields is called Lorentz force. It is given by

$\text{F}=q\left(\text{E}+\text{v}\times \text{B}\right)$

Motion of a Charged Particle in a Uniform Magnetic Field

When charged particle enter normal to the magnetic field it follows a circular path.

The radius of the path, $r=\frac{mv}{Bq}$

$\therefore$ $r\propto mv$

and $r\propto \frac{1}{\left(q/m\right)}$

Time period, $T=\frac{2\pi m}{Bq}$

When charged particle enter magnetic field at any angle except $90{}^{\text{o}},$ then it follows helical path.

The radius of the path, $r=\frac{mv\sin \theta }{Bq}$

The distance travelled by the charged particle in one time period due to component of velocity $v\cos \theta ,$ is called pitch of the path.

$Pitch=T\times v\cos \theta =\frac{2\pi mv\cos \theta }{Bq}$