TORQUE ON A CURRENT LOOP IN A UNIFORM MAGNETIC FIELD
We have seen that the net magnetic force on a closed loop carrying a current is zero when the loop is placed in an external uniform magnetic field. Although this is generally true, there is a net torque on a current loop in an external field, and this torque tends to rotate the loop.
To understand this, first consider a rectangular loop carrying a current/ in the presence of a uniform magnetic field in the plane of the loop as in Fig. [5.23 (a)]. The forces on the sides of length b are zero since these
wires are parallel to the field and hence dl x B=0 for these sides. On the other hand, the magnitude of the forces on the sides of the length I is given
where F\ is the force on the left side of the loop and Fi is the force on the right side.
The direction jf F\ is out of the paper in —»
Fig. (5.23) and that of Fi is into the paper. An end viev/ of the direction of these forces, shown in Fig. 5.23 (b), demonstrates that the forces form a couple. Therefore, even though they cancel each other, they tend to rotate the loop. The torque due to this couple about the point 0 in Fig. [5.23 (6)]
has a magnitude given by
T = Fi- + F2– = IIbB . 2 2
where the moment arm about 0 is b/2 for both forces.This in fact, is the torque about any point. But the area of the loop is given by A = ab. Hence, the torque can be expressed as
T=IAB | … (0
Note that this result is valid only when the field B is in the plane of the loop. The sense pf the rotation is clockwise when viewed from the bottom end, as indicated in Fig. 5.23 (b). If the current were reversed, the forces would reveise their directions and the rotational tendency would be counterclockwise.
Now consider a rectangular loop carrying a current I in a uniform magnetic field. Suppose the field makes an angle 9 with the normal to the plane of the loop as in Fig. 5.24 (a). For convenience, we shall assume that —«
the field B is perpendicular to the sides of length I. In this case, the magnetic forces Fi and Fa on the sides of length b cancel each other and produce no torque since they pass through a common origin 0. However, the forces F,and F2 acting on the sides of length I form a couple and hence produce a torque about any point 0. Referring to the end view shown in Fig. 5.24 (b) we note that the moment arm of the force F\ about the point 0 is equal to — sin 9. Likewise, the moment arm of Fj about 0 is also — sin 0. 2 2
Since Fl = F2 = / / B, the net torque about 0 has a magnitude given by
T = Fi – sin 0 + F2- sin 0 .2 2
= I IB || sin flj + I IB sin ©j = IlbB sin 9
= IAB sin 9
where A = Ibis the area of the loop. This result shows that the torque has the maximum value IAB when the field is parallel to the plane of the loop (9 = 90°) and is zero when the field is perpendicular to the plane of the loop (9 = 0). As we see in Fig. (5.24) the loop tends to rotate to smaller values of 9 (that is, such that the normal to the plane of the loop rotates 1
A convenient vector expression for the torque is the following cross product relationship:
T = /AxB …(/0
where A , a vector perpendicular to the plane of the loop, has a magnitude equal
to the area of the loop. The sense of A is determined by the right-hand rule as described in Fig. (5.25). By wrapping the four fingers of the right hand in the direction of the current, the thumb points in the direction of A. The product IA is defined to the magnetic —# moment m of the loop. That is,
‘ —* —t
m = IA … (iii)
The SI unit of magnetic moment is ampere metre2 (A.nt). Using this definition the torque can be expressed
r = mxB … (iv)
Magnetic moment of the coil is usually defined as the couple exerted on the coil when placed with its plane parallel to a field of unit magnitude.
Note that this result is analogous to the torque acting on an electric dipole moment p in the presence of an external electric field £ where
T = p x £. If a coil has N turns all of the same dimensions, the magnetic moment and the torque on the coil will clearly be N times greater than in single loop.
Although the torque was obtained for a particular orientation of B with respect to the loop the equation r = mxB is valid for any orientation.Furthermore, although the torque expression was derived for a rectangular loop, the result is valid for a planner loop of any shape.
Potential Energy of A Current Loop In A Magnetic Field
Let a current loop be placed in a magnetic field B and at any instant let the normal to its plane make an angle 0 with the magnetic field B.
Then, the torque on the current loop is given, by .
—» —i T = IA xB
or 1 = IAB sin 0 (magnitude)
Let the current loop be turned further through an angle dQ. The work done is given by
dW = xd9 = IAB sin QdQ Now, the total work done in turning the current loop (dipole) through an angle 0 is given by
IdW = 1 IAB sin0 dQ This by definition is. called the potential energy of the current loop (dipole) in this orientation. It is represented by U.
:. Potential energy of a current loop (dipole) = U = J IAB isin 6 dQ or U = -IAB cos Q +c where c is constant of integration.
Let us take U = 0 when & = — , then c = o
U -—IAB cos 6 =-mB cos 0 —• —i —.
or U =-m.B where m is the magnetic moment of the loop.
This corresponds to the relation -» —*
U =-p.E for the potential energy of a dipole in electric field.
It is interesting to note the similarity between the rotating tendency of a current loop in an external magnetic field and the motion of a compass needle (or pivoted bar magnet) in such a field. Like the current loop, the compass needle and bar magnet can be regarded as magnetic dipoles. The similarity in their magnetic field lines is described in Fig.5.26. Note that one face of the current loop behaves as the north pole of a bar magnet while the opposite face behaves as the south pole. The field lines shown in fig. 5.26 are the patterns due to the bar magnet [Fig.5.26(a)J and the current loop [Fig.5.26(b)]. There is no external field present in these diagrams. Furthermore, the diagrams are a simplified, two-dimensional description of . the field lines.
Calculate the couple needed to hold a coil of area 8 cm2 at an angle of 60° to a fixed of flux density 0-1 T if the coil carries current of 0-5A. Solution
Couple =BA1 sin 9 = 0-1 x (8 x 10″4) x 0-5 sin 60° = 0-1 x (8 x 10^) x 0-5 x 0-866 Nm
=| 346x lQ-sAfoT