Torque on a rectangular current loop in a uniform magnetic field
`color{blue} ✍️` We now show that a rectangular loop carrying a steady current `I` and placed in a uniform magnetic field experiences a torque.
`color {blue}{➢➢}`It does not experience a net force. This behaviour is analogous to that of electric dipole in a uniform electric field (Section 1.10). We first consider the simple case when the rectangular loop is placed such that the uniform magnetic field `B` is in the plane of the loop.
`color {blue}{➢➢}`This is illustrated in Fig. 4.21(a). The field exerts no force on the two arms `AD` and `BC` of the loop. It is perpendicular to the arm `AB` of the loop and exerts a force `F_1` on it which is directed into the plane of the loop. Its magnitude is,
`color{blue}(F_1 = IbB)`
`color {blue}{➢➢}`Similarly it exerts a force `F_2` on the arm `CD` and `F_2` is directed out of the plane of the paper.
`color{blue}(F_2 = I b B = F_1)`
`color {blue}{➢➢}`Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces `F_1` and `F_2.` Figure 4.21(b) shows a view of the loop from the `AD` end. It shows that the torque on the loop tends to rotate it anti-clockwise. This torque is (in magnitude).
`color{blue}(tau = F_1 a/2 + F_2a/2)`
`color{blue}(= IbB a/2 + I bB a/2 = I(ab)B)`
`color {blue}{➢➢}`where `A = ab` is the area of the rectangle.
We next consider the case when the plane of the loop, is not along the magnetic field, but makes an angle with it. We take the angle between the field and the normal to the coil to be angle `θ` (The previous case
corresponds to `θ = π/2`).
`color{blue} ✍️` Figure 4.22 illustrates this general case. The forces on the arms `BC` and `DA` are equal, opposite, and act along the axis of the coil, which connects the centres of mass of `BC` and `DA`.
`color {blue}{➢➢}` Being collinear along the axis they cancel each other, resulting in no net force or torque. The forces on arms `AB` and `CD` are `F_1` and `F_2`. They too are equal and opposite, with magnitude, `F_1 = F_2 = I bB` But they are not collinear! This results in a couple as before.
`color{blue} ✍️` The torque is, however, less than the earlier case when plane of loop was along the magnetic field. This is because the perpendicular distance between the forces of the couple has decreased. Figure 4.22(b) is a view of the arrangement from the AD end and it illustrates these two forces constituting a couple. The magnitude of the torque on the loop is,
`color{blue}(tau = F_1 a/2 sin theta + F_2 a/2 sin theta)`
`color{blue}(=I ab B sin θ)`
`color {blue}{➢➢}` As `color{blue}(θ -> 0,)` the perpendicular distance between the forces of the couple also approaches zero. This makes the forces collinear and the net force and torque zero.
`color{blue} ✍️`The torques in Eqs. (4.26) and (4.27) can be expressed as vector product of the magnetic moment of the coil and the magnetic field. We define the magnetic moment of the current loop as,
`color {blue}{➢➢}` where the direction of the area vector `A` is given by the right-hand thumb rule and is directed into the plane of the paper in Fig. 4.21. Then as the angle between `m` and B is `θ` , Eqs. (4.26) and (4.27) can be expressed by one expression
`color{blue} ✍️` This is analogous to the electrostatic case (Electric dipole of dipole moment `p_e` in an electric field `E`).
`color{blue}(tau = P_e xx E)`
`color {blue}{➢➢}` As is clear from Eq. (4.28), the dimensions of the magnetic moment are `[A][L^2]` and its unit is `Am^2.` From Eq. (4.29), we see that the torque `τ` vanishes when `m` is either parallel or antiparallel to the magnetic field `B`.
`color {blue}{➢➢}` This indicates a state of equilibrium as there is no torque on the coil (this also applies to any object with a magnetic moment `m`). When `m` and `B` are parallel the equilibrium is a stable one. Any small rotation of the coil produces a torque which brings it back to its original position.
`color {blue}{➢➢}` When they are antiparallel, the equilibrium is unstable as any rotation produces a torque which increases with the amount of rotation. The presence of this torque is also the reason why a small magnet or any magnetic dipole aligns itself with the external magnetic field. If the loop has `N` closely wound turns, the expression for torque, Eq. (4.29), still holds, with.
`color {blue}{➢➢}`It does not experience a net force. This behaviour is analogous to that of electric dipole in a uniform electric field (Section 1.10). We first consider the simple case when the rectangular loop is placed such that the uniform magnetic field `B` is in the plane of the loop.
`color {blue}{➢➢}`This is illustrated in Fig. 4.21(a). The field exerts no force on the two arms `AD` and `BC` of the loop. It is perpendicular to the arm `AB` of the loop and exerts a force `F_1` on it which is directed into the plane of the loop. Its magnitude is,
`color{blue}(F_1 = IbB)`
`color {blue}{➢➢}`Similarly it exerts a force `F_2` on the arm `CD` and `F_2` is directed out of the plane of the paper.
`color{blue}(F_2 = I b B = F_1)`
`color {blue}{➢➢}`Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces `F_1` and `F_2.` Figure 4.21(b) shows a view of the loop from the `AD` end. It shows that the torque on the loop tends to rotate it anti-clockwise. This torque is (in magnitude).
`color{blue}(tau = F_1 a/2 + F_2a/2)`
`color{blue}(= IbB a/2 + I bB a/2 = I(ab)B)`
` color{blue}(= IAB)`
..........(4.26)`color {blue}{➢➢}`where `A = ab` is the area of the rectangle.
We next consider the case when the plane of the loop, is not along the magnetic field, but makes an angle with it. We take the angle between the field and the normal to the coil to be angle `θ` (The previous casecorresponds to `θ = π/2`).
`color{blue} ✍️` Figure 4.22 illustrates this general case. The forces on the arms `BC` and `DA` are equal, opposite, and act along the axis of the coil, which connects the centres of mass of `BC` and `DA`.
`color {blue}{➢➢}` Being collinear along the axis they cancel each other, resulting in no net force or torque. The forces on arms `AB` and `CD` are `F_1` and `F_2`. They too are equal and opposite, with magnitude, `F_1 = F_2 = I bB` But they are not collinear! This results in a couple as before.
`color{blue} ✍️` The torque is, however, less than the earlier case when plane of loop was along the magnetic field. This is because the perpendicular distance between the forces of the couple has decreased. Figure 4.22(b) is a view of the arrangement from the AD end and it illustrates these two forces constituting a couple. The magnitude of the torque on the loop is,
`color{blue}(tau = F_1 a/2 sin theta + F_2 a/2 sin theta)`
`color{blue}(=I ab B sin θ)`
`color{blue}(= I A B sin θ)`
.........(4.27)`color {blue}{➢➢}` As `color{blue}(θ -> 0,)` the perpendicular distance between the forces of the couple also approaches zero. This makes the forces collinear and the net force and torque zero.
`color{blue} ✍️`The torques in Eqs. (4.26) and (4.27) can be expressed as vector product of the magnetic moment of the coil and the magnetic field. We define the magnetic moment of the current loop as,
`color{blue}{m = I A}`
..............(4.28)`color {blue}{➢➢}` where the direction of the area vector `A` is given by the right-hand thumb rule and is directed into the plane of the paper in Fig. 4.21. Then as the angle between `m` and B is `θ` , Eqs. (4.26) and (4.27) can be expressed by one expression
`color{blue}{τ = m × B}`
............(4.29)`color{blue} ✍️` This is analogous to the electrostatic case (Electric dipole of dipole moment `p_e` in an electric field `E`).
`color{blue}(tau = P_e xx E)`
`color {blue}{➢➢}` As is clear from Eq. (4.28), the dimensions of the magnetic moment are `[A][L^2]` and its unit is `Am^2.` From Eq. (4.29), we see that the torque `τ` vanishes when `m` is either parallel or antiparallel to the magnetic field `B`.
`color {blue}{➢➢}` This indicates a state of equilibrium as there is no torque on the coil (this also applies to any object with a magnetic moment `m`). When `m` and `B` are parallel the equilibrium is a stable one. Any small rotation of the coil produces a torque which brings it back to its original position.
`color {blue}{➢➢}` When they are antiparallel, the equilibrium is unstable as any rotation produces a torque which increases with the amount of rotation. The presence of this torque is also the reason why a small magnet or any magnetic dipole aligns itself with the external magnetic field. If the loop has `N` closely wound turns, the expression for torque, Eq. (4.29), still holds, with.
`color{blue}(m = N I A)`
...............(4.30)
