TORQUE ON A CURRENT CARRYING LOOP
Consider a rectangular current carrying loop PQRS carrying a current i as shown in the figure below. The magnetic forces `F_1, F_2, F_3 and F_4` on the wires `PQ, QR, RS` and `SP` respectively are obtained from equation `vecF = ivecl xx vecB`. These forces act from the middle point `T, U, V, W` of the respective sides. Hence `F_1 = F_3 = ilB` and `F_2 = F_4 = ibB`. The resultant force is therefore zero.
Also, `F_1` and `F_3` have the same line of action so they together produce no torque. Similarly `F_2` and `F_4` produce no torque.
Now suppose the loop is rotated through an angle `theta` about the line WU. The wire PQ shifts parallel to itself so that the force `vecF_1 = ivecl xx vecB` on it remains unchanged in magnitude and direction. Its point of application T shifts toT'. Similarly, the force on RS remains `F_3`. But the point of application shifts to V. The line TV gets rotated by an angle `theta` to take the position T' V'. This line makes an angle `theta` with the force `F_1` and `F_3`. The torque of `F_1` about O has magnitude.
`|OT' xx F_1| = (b/2) xx F_1 sintheta = b/2(ilB) sintheta`
This torque acts along the line UW. The torque of `F_3` about 0 is also `b/2(ilB) sintheta`
Along the same direction. As the wire QR rotates about WU, the plane containing the wire and the ge. The force on this wire is perpendicular to this plane and hence its direction remains unchanged. Also, the point of application U remains the same. Similar in the case for the wire SP. The forces on QR and SP are, therefore, equal and opposite and act along the same line. They together produce no torque. The net torque acting on the loop is therefore,
`T = b/2(ilB) sintheta +b/2(ilB) sintheta = b(ilB) sintheta`
`= iAB sintheta`
Where A is the magnitude of area of the loop PQRS. The direction of the area vector `vecA` Is perpendicular to the plane of the loop and is towards the side from which the current looks anti- clockwise. Thus in the figure the area vector points towards a viewer. It is drawn from the center O of the loop. Another way to get the direction of `vecA` is to use the righthand thumb rule. If you curl your fingers of the right hand along the current, the stretched thumb gives the direction of `vecA` In the figure (a) the angle between the area vector `vecA` and the magnetic field is zero. As the loop rotates the areavector also rotates by an angle `theta`. Taking the direction of torque (along UWJ into consideration,
`vectau = ivecA xx vecB = vecmu xx vecB` ...................(1)
where `vecmu = ivecA` is called the magnetic dipole moment or simply magnetic moment of the current loop.We know that in a electric dipole when the pair ofcharges +q and -q are separated by a distance I then the electric dipole moment is given by p = ql and
in the direction from -q to +q. If such a dipole is placed in a uniform electric field, a torque `vectau = vecp xx vecE` acts on the dipole.
Equation 1 is similar to this structure and hence `vecmu` is called the magnetic dipole moment. For n number ofturns we write magnetic dipole moment as
`vecmu = nivecA` ..........................................(2)
The equations 1 and 2 obtained are shown for a rectangular loop but are also valid for a plane loop of any shape. Every current carrying loop can be treated as a bar magnet. In a bar magnet, magnetic fileld lines emanatesfrom the north pole and after forming
a closed path terminates on south pole. Similarly a current carrying loop can be imagined to have a north pole and South Pole or in simple words it can be imagined that a bar magnet is placed inside tlle loop. Thus current carrying loop forms a magnetic dipole. Now we know that the direction of magnetic moment is from south pole to north pole and that the direction of magnetic moment of
the loop is in the direction of area vector `vecA` as in Eqn. 2. Since the direction of `vecA` is decided by the stretched thumb as discussed above the stretched thumb will point in the direction or North Pole.
Also, `F_1` and `F_3` have the same line of action so they together produce no torque. Similarly `F_2` and `F_4` produce no torque.
Now suppose the loop is rotated through an angle `theta` about the line WU. The wire PQ shifts parallel to itself so that the force `vecF_1 = ivecl xx vecB` on it remains unchanged in magnitude and direction. Its point of application T shifts toT'. Similarly, the force on RS remains `F_3`. But the point of application shifts to V. The line TV gets rotated by an angle `theta` to take the position T' V'. This line makes an angle `theta` with the force `F_1` and `F_3`. The torque of `F_1` about O has magnitude.
`|OT' xx F_1| = (b/2) xx F_1 sintheta = b/2(ilB) sintheta`
This torque acts along the line UW. The torque of `F_3` about 0 is also `b/2(ilB) sintheta`
Along the same direction. As the wire QR rotates about WU, the plane containing the wire and the ge. The force on this wire is perpendicular to this plane and hence its direction remains unchanged. Also, the point of application U remains the same. Similar in the case for the wire SP. The forces on QR and SP are, therefore, equal and opposite and act along the same line. They together produce no torque. The net torque acting on the loop is therefore,
`T = b/2(ilB) sintheta +b/2(ilB) sintheta = b(ilB) sintheta`
`= iAB sintheta`
Where A is the magnitude of area of the loop PQRS. The direction of the area vector `vecA` Is perpendicular to the plane of the loop and is towards the side from which the current looks anti- clockwise. Thus in the figure the area vector points towards a viewer. It is drawn from the center O of the loop. Another way to get the direction of `vecA` is to use the righthand thumb rule. If you curl your fingers of the right hand along the current, the stretched thumb gives the direction of `vecA` In the figure (a) the angle between the area vector `vecA` and the magnetic field is zero. As the loop rotates the areavector also rotates by an angle `theta`. Taking the direction of torque (along UWJ into consideration,
`vectau = ivecA xx vecB = vecmu xx vecB` ...................(1)
where `vecmu = ivecA` is called the magnetic dipole moment or simply magnetic moment of the current loop.We know that in a electric dipole when the pair ofcharges +q and -q are separated by a distance I then the electric dipole moment is given by p = ql and
in the direction from -q to +q. If such a dipole is placed in a uniform electric field, a torque `vectau = vecp xx vecE` acts on the dipole.
Equation 1 is similar to this structure and hence `vecmu` is called the magnetic dipole moment. For n number ofturns we write magnetic dipole moment as
`vecmu = nivecA` ..........................................(2)
The equations 1 and 2 obtained are shown for a rectangular loop but are also valid for a plane loop of any shape. Every current carrying loop can be treated as a bar magnet. In a bar magnet, magnetic fileld lines emanatesfrom the north pole and after forming
a closed path terminates on south pole. Similarly a current carrying loop can be imagined to have a north pole and South Pole or in simple words it can be imagined that a bar magnet is placed inside tlle loop. Thus current carrying loop forms a magnetic dipole. Now we know that the direction of magnetic moment is from south pole to north pole and that the direction of magnetic moment of
the loop is in the direction of area vector `vecA` as in Eqn. 2. Since the direction of `vecA` is decided by the stretched thumb as discussed above the stretched thumb will point in the direction or North Pole.