# Force and Torque on Conductors

**Force and Torque on Conductors :**

**Force on a conductor:**

Ampere proposed that if a conductor carrying current produces a magnetic field and exerts a force on a magnet, then the magnet should also exert a force on the conductor carrying current. Examples: - Suspend an aluminum rod horizontally on the wire between the poles of a horseshoe magnet, and pass an electric current through the wire, the aluminum rod will be displaced. If the direction of the current is reversed, the direction of displacement will also be reversed. The applied force is greatest when the conductor is perpendicular to the magnetic field.

Fleming's Left Hand Rule: –

The direction of the force (movement) of a current-carrying conductor in a magnetic field is determined by Fleming's left-hand rule. "If the thumb, index and middle fingers of the left hand are at right angles to each other so that the index finger points in the direction of the magnetic field and the middle finger points in the direction of the electric current, the thumb points in the direction of the force (direction of movement) of the conductor."

**Force on a Conductor Carrying Current in a Magnetic field:**

When a current-carrying conductor is placed in a uniform magnetic field, a force is exerted on the moving charges inside the conductor (due to the Lorentz force). This is the force required in a conductor through which current flows. To calculate this force, we consider various parameters as follows:

L is the conductor length,

i am the current that flows in him

q is the charge flowing through the conductor at time 't'.

v is the velocity of charge q,

B is the uniform magnetic field in which there is a conductor carrying current.

**A special case of force on a live conductor **

There are some special cases where a force acts on a current-carrying conductor, depending on the position of the conductor within the magnetic field. These cases are described as follows.

Case I: Conductor placed parallel to the magnetic field

When sin θ = 0 (minimum), i. H. θ = 0° or 180°, the force on the current element in the magnetic field is zero (minimum).

F_{min} = 0

If the current in the magnetic field is collinear with the magnetic field, the current element in the magnetic field experiences no force. Therefore, the force experienced by a conductor within a given magnetic field is minimal.

Case II: Conductor placed perpendicular to the magnetic field

For sin θ = 1 (maximum), i.H. θ = 90°, the force on the current element is maximum in the magnetic field (=ILB).

Fmax = ILB

The force direction is always perpendicular to the containing plane. L and B, which is the maximum force a conductor can experience in a given magnetic field.

**Force Between two Parallel Current-Carrying Conductors :**

We learned about the existence of magnetic fields caused by current-carrying conductors and the Biot-Savart law.

We also learned that an external magnetic field exerts a force on a current-carrying conductor and the Lorentz force formula that underlies this principle.

Therefore, two studies show that two conductors carrying current exert a magnetic force on each other when they are placed close to each other. This section details this case.

Consider the system shown in the diagram above. Here we have two parallel current conductors separated by a distance 'd' such that one conducts the current I1 and the other conducts the current I2, as shown in the figure. Based on what we just learned, we can say that Conductor 2 experiences the same magnetic field at any point along its length by conductor 1. The direction of magnetic force is shown in the diagram and can be found using the right-hand thumb rule. As you can see, the direction of the magnetic field is down due to the first conductor. From Ampere's circuit law, the magnitude of the magnetic field due to the first conductor is given by

The force exerted by conductor 1 on a segment of length L of conductor 2 is given by

Similarly, we can calculate the force that conductor 2 exerts on conductor 1. It can be seen that conductor 1 is subjected to the same force by conductor 2 but in the opposite direction. therefore,

It has also been observed that currents flowing in the same direction attract conductors to each other, and currents in opposite directions cause conductors to repel each other. The magnitude of the force acting per unit length is given by

**Torque on a Conductor:**

**Torque on Current Carrying Coil in Magnetic Field:**

A magnetic dipole is the limit of either a closed electric circuit or a pair of poles because the size of the source is reduced to zero keeping the magnetic moment constant. It is now shown that a constant current I through a rectangular loop placed in a uniform field has torque. It has no mains power. This behavior is similar to an electric dipole in a uniform electric field.

Consider the case of placing a rectangular loop such that the uniform magnetic field B is in the plane of the loop. The field exerts no force on her two arms PS and QR in the loop. It is perpendicular to arm PQ of the loop and exerts a force F1 on arm PQ directed in the plane of the loop. Its size is

F_{1}=IzB

Similarly, force F_{2} is applied to arm RS and F_{2 }is directed out of the plane of the paper.

F_{2} =IzB= F_{1 }

Therefore the net force on the loop is zero. A torque acts on the loop because the two forces F_{1} and F_{2 }cancel each other. Here we can see that the torque applied to the loop tends to cause it to rotate counterclockwise.

τ = F_{1} $\left(\frac{Y}{2}\right)$ + F_{2} $\left(\frac{Y}{2}\right)$

= Izb$\left(\frac{Y}{2}\right)$ + Izb$\left(\frac{Y}{2}\right)$

= I(yz) B

= IAB ……. (1)

Where A = y × z is the area of the rectangle.

** Case 2 **

Now consider the case where the plane of the loop is not aligned with the magnetic field, but at an angle with it. And consider the angle between the magnetic field and the normal to the coil as the angle Θ. The forces on the two arms, QR and SP, are equal and opposite, acting along the axis of the coil connecting the centers of mass of QR and SP. Since they are collinear along the axis, they cancel each other out and produce no net force or torque. The forces on arms PQ and RS are F_{1 }and F_{2}. Moreover, they are also of the same and opposite magnitude,

F_{1} = F_{2} = IzB

Since they are not collinear, they form a pair. However, the effect of torque is less than if the plane of the loop passed along the magnetic field. The magnitude of the torque on the loop is:

τ = F_{1} $\left(\frac{Y}{2}\right)$ sinθ F_{2 $\left(\frac{Y}{2}\right)$ }inθ

= I(y×z) B sin θ

= IABsinθ ……. (2)

Therefore, the torque in equations (1) and (2) can be expressed as the vector product of the magnetic moment of the coil and the magnetic field. Therefore, the magnetic moment of the current loop can be defined as

m = IA

where A is the direction of the area vector. If the angle between m and B is θ, equations (1) and (2) can be expressed as follows.

τ = m × B

where m is the magnetic moment and B is the uniform magnetic field.

**Magnetic Dipole Moment:**

The magnetic moment is a quantity that describes the magnetic strength and orientation of a magnet or other object that produces a magnetic field. More specifically, magnetic moment refers to the magnetic dipole moment, i.e. the component of the magnetic moment that can be represented by a magnetic dipole. A magnetic dipole is a magnetic north pole and a magnetic south pole separated by a small distance.

The magnetic dipole moment is the current and area or energy divided by the magnetic flux density. The dipole moment has units of meters, kilograms, seconds, amperes, and amperes squared. The unit of the centimeter-gram-second system is erg (unit of energy)/gauss (unit of magnetic flux density). 1,000 ergs per gauss equals 1-ampere square meter.

**Magnetic Moment formula:**

Magnetic dipole moment – The magnetic field B due to a current loop carrying current i of radius R separated by a distance l along the axis is given by

Now, considering points so far from the current loop that l>>R holds, we can approximate the field as

Now the area of the loop is A

So the magnetic field can be written as

We can write this new quantity μ as a vector pointing along the magnetic field.

Notice the striking resemblance to the electric dipole field.

Unlike the electric field, the magnetic field has no "charge" or "counterpart". In other words, the magnetic field has no source or sink, only dipoles can exist. Anything that can generate a magnetic field has both a source and a sink. H. There are both north and south poles. In many respects, magnetic dipoles are fundamental entities capable of creating magnetic fields.

Most elementary particles behave essentially like magnetic dipoles. For example, the electron itself behaves like a magnetic dipole and has a magnetic spin dipole moment. This magnetic moment is inherent because the electron has no region A (it is a point body) and does not rotate around itself, but it is fundamental to the nature of the electron's existence.

The 'N' times magnetic moment of a wire loop can be generalized as

μ = NiA

The field lines of the current loop resemble those of an ideal electric dipole.

If you've ever split a magnet in two, you'll know that each piece forms a new magnet. The new pieces also include the north and south poles. It seems impossible to get just the North Pole.

**Current Loop as a Magnetic Dipole:-**

Ampere found that the distribution of magnetic lines of force around a finite current-carrying solenoid is similar to that produced by a bar magnet. This is evident from the fact that a compass needle when similar deflections moved around these two bodies. After noting the close resemblance between these two, Ampere demonstrated that a simple current loop behaves like a bar magnet and put forward that all the magnetic phenomena are due to circulating electric current. This is Ampere’s hypothesis.

The magnetic induction at a point along the axis of a circular coil carrying current is,

B = µ_{o}nI$\frac{{a}^{2}}{2}$(a^{2}+x^{2})$\frac{3}{2}$

The direction of this magnetic field is along the axis and is given by the right-hand rule. For points that are far away from the center of the coil, x>>a, a^{2} is small and it is neglected. Hence for such points,

B = µ_{o}nI$\frac{{a}^{2}}{2}$x3

If we consider a circular loop, n = 1, its area A = πa^{2}

_{o }I$\frac{{a}^{2}}{2}$πx3

The magnetic induction at a point along the axial line of a short bar magnet is

B = $\left(\frac{{\mu}_{o}}{4\pi}\right)\left(\frac{2M}{x3}\right)$

Or, B = $\left(\frac{{\mu}_{o}}{2\pi}\right)\left(\frac{M}{x3}\right)$

Comparing equations (1) and (2), we find that

M = IA

Hence a current loop is equivalent to a magnetic dipole of moment M = IA

The magnetic moment of a current loop is defined as the product of the current and the loop area. Its direction is perpendicular to the plane of the loop.

**Magnetic Dipole Moment of a Spinning Electron:**

According to Neil Bohr's atomic model, a negatively charged electron revolves around a positively charged nucleus in a circular orbit of radius r. An electron rotating in a closed path generates an electric current. The counterclockwise movement of the electron produces a clockwise current in the conventional current.

Since the electron is negatively charged, the normal current flows in the opposite direction to its movement. The magnetic moments associated with the orbital motion of two electrons are shown in the figure. An electron rotating in an orbit of radius 'r' corresponds to a magnetic layer with a magnetic moment of 'M'

M = i\ times A

Here "i" is the current corresponding to the spinning electron and A is the area of the orbit. Where '\tau' is the spin time of the electron.

i = charge/period = e/\tau = $\frac{e}{{\displaystyle \frac{2\pi}{\omega}}}$

or i = $\frac{e\omega}{2\pi}$

Here 'ω' is the angular velocity of the electron.

So M = $\frac{e\omega}{2\pi}$\ times (πr^{2}) = $\frac{e\omega {r}^{2}}{2}$

According to Bohr's theory, an electron can move in an orbit where its angular momentum is an integral multiple of h/2π, where "h" is Planck's constant.

mr^{2}ω = n $\left(\frac{h}{2\mathrm{\pi}}\right)$

Or r^{2}ω = n $\left(\frac{h}{2m\mathrm{\pi}}\right)$

Substituting r^{2}ω in the equation, we get,

M = n $\left(\frac{eh}{\mathrm{\pi m}}\right)$

Since an atom can have a large number of electrons, the magnetic moment of an atom is the vector sum of the magnetic moments of the different electrons. The term $\left(\frac{eh}{\mathrm{\pi m}}\right)$ is called the Bohr magneton. This is the smallest value of the magnetic moment of the electron. Each atom can have a magnetic moment that is a fixed multiple of the Bohr magneton. Thus, the magnetic moment is also quantified at the atomic and subatomic level.

In addition to the magnetic moment due to orbital motion, the electron has a magnetic moment due to its spin. Thus, the resultant magnetic moment of an electron is the vector sum of its orbital magnetic moment and its spin magnetic moment.