Presentation on theme: "Torque On A Current Loop In A Uniform Magnetic Field"— Presentation transcript:
1Torque On A Current Loop In A Uniform Magnetic Field AP Physics CMontwood High SchoolR. Casao
2With the knowledge that a force is exerted on a current-carrying conductor when the conductor is placed in an external magnetic field, this force can produce a torque on a current loop placed in the magnetic field.Consider a rectangular loop carrying a current I in the presence of a uniform magnetic field in the plane of the loop.
3The forces on the sides of length b are zero since these wires are parallel to the field (ds X B = 0 for these sides).The magnitude of the forces on the sides of length a is given by:F2 = F4 = I·a·BThe direction of F2 is out of the page; the direction of F4 is into the page.
4If viewing the loop from the end, and we assume that the loop is pivoted so that it can rotate about point O, the two forces produce a torque about O that rotates the loop clockwise.The magnitude of the torque, which is maximum in this position, is:Tmax = I·a·b·B:
5The area of the loop A = a·b; Tmax = I·A·B If the current direction were reversed, the forces would reverse their directions and the rotational tendency would be counterclockwise.Suppose the magnetic field makes an angle q with respect to a line perpendicular to the plane of the loop (the dashed line).
6The vector A which is perpendicular to the plane of the loop is the area vector A of the loop. Only the forces F2 and F4 contribute to the torque about the axis of rotation O. The other two forces on the loop would not produce a rotation as these forces would be equal in magnitude and opposite in direction and would also pass through the axis of rotation O, making the torque arm 0 m.
8The forces F2 and F4 form a couple and produce a torque about any point. We will use point O as the pivot point and I’m going to resolve the forces in to a component that is parallel to the axis of the loop and a component that is perpendicular to the axis of the loop.
9The parallel component of the force Fp will not produce a rotation because it passes through the pivot point.The perpendicular component of the force F will produce the torque that will cause the loop to rotate.
10Net torque about point O: T = 2·F2·sin ·b/2 T = F2·b·sin Remember: For each perpendicular force:r = b/2F = F2·sin T = F2·sin ·b/2Net torque about point O:T = 2·F2·sin ·b/2T = F2·b·sin Remember:F2 = I·a·BT = I·a·B·b·sin T = I·A·B·sin
11The torque has a maximum value I·A·B when the magnetic field is parallel to the plane of the loop (angle q between A and B = 90°).The torque is 0 N·m when the magnetic field is perpendicular to the plane of the loop (angle q between A and B = 0° = 180°).The loop tends to rotate to smaller values of q (so that the area vector A rotates aligns with the magnetic field vector B).
12To express the torque as a vector cross product: The direction of the area vector A is determined by the right hand rule: rotate the fingers of the right hand in the direction of the current in the loop, the thumb points in the direction of the area vector A.The direction of the torque is also given by the right hand rule: point the fingers of the right hand in the direction of the area vector A, the palm points in the direction of the magnetic field B, the thumb points in the direction of the torque.The product I·A is defined as the magnetic moment m of the loop; m = I·A
13T = N·I·A·B = N·(m X B)=N·m·B·sin q Unit for magnetic moment m = A·m2.T = m x B (for a single loop)This is the same type of torque that acts on an electric dipole moment p in an external electric field to align the dipole moment with the electric field E; T = p X EIf a coil in a magnetic field B has N loops of the same dimensions, the magnetic moment and the torque on the coil will be N times greater than a single loop.T = N·I·A·B = N·(m X B)=N·m·B·sin q
14The Plane of the Loop Most problems refer to the plane of the loop. The plane of the loop is the plane that lies along a straight line between the opposing currents in the conducting loop.The plane of the loop is perpendicular to the area vector.
15The Plane of the LoopMaximum torque occurs when the plane of the loop is parallel to the magnetic field B.Angle between plane of loop and B is 0°.Angle between area vector A and B is 90°.Zero torque occurs when the plane of the loop is perpendicular to the magnetic field B.Angle between plane of loop and B is 90°.Angle between area vector A and B is 0°.
16Work and Torque Work = Torque·angular displacement q Generally, the angular displacement q is measured in radians.For a displacement from qi to qf:
17Potential Energy and Torque The potential energy U of a system of a magnetic dipole in a magnetic field depends on the orientation of the dipole in the magnetic field.The system has its lowest U when m points in the direction of B; the angle between m and B = 0°.The system has its highest U when m points in the opposite direction of B; the angle between m and B = 180°.