Physics II Magnetostatics

Why should you care? Practical uses of magnetism: Magnetic resonance imaging (MRI) Magnetic hypothermia Magneto-optic screening A neurosurgeon can: Guide a pellet through brain tissue to inoperable tumors. Pull a catheter into place. Implant electrodes while doing little harm to brain tissue.

Magnetostatics Electrostatics: All sources charges are static, meaning they aren’t moving.. Produces an electric field 𝐸 . Magnetostatics: All charges are moving, but the movement is static, meaning it’s a constant rate of movement. Produces a magnetic field 𝐵 .

Observational experiment Imagine the following: Two wires hand from the ceiling, a few centimeters apart. I turn on a current. Do the wires… Repel each other? Attract each other? Remain where they are?

Observational experiment a) Currents in opposite directions repel. b) Currents in same directions attract.

Take-away This is our first encounter with the magnetic force. Stationary charge produces only the electric field 𝐸 . Moving charge, in addition, produces the magnetic field 𝐵 .

Magnetic fields As with the electric field, we can represent the magnetic field by means of drawings with magnetic field lines. These field lines always go from north to south. In a magnet, as shown here, charge seems to be stationary, but that is not the case. If you could examine a piece of magnetic material on an atom scale, you would find tiny currents: electrons orbiting around nuclei and electrons spinning on their axes.

Magnetic fields Our earth has magnetic north and south. The field protects use from cosmic particles.

Magnetic fields If you hold up a compass outside, the needle will point to the north. But if you hold a compass up to a current-carrying wire, the result is peculiar. The compass will not point toward the wire, nor away from it, but rather it circles around the wire.

Magnetic fields How can such a field create an attraction between two wires? On the second wire… The magnetic field is into the page. The velocity of the charges is upward. The resulting force is to the left. A very strange law is needed to account for these directions.

Magnetic force If you’re perceptive, you’ll notice the odd behavior follows what is knows as the Right-Hand Rule.

Magnetic force Yes, we’re speaking in three dimensions now.

Magnetic force 𝐹 𝐵 =𝑞 𝑣 × 𝐵 𝐹 𝐵 =𝑞𝑣𝐵sin𝜃 Mathematically, this is described by a cross product. 𝐹 𝐵 =𝑞 𝑣 × 𝐵 A cross product simply says that 𝐹 𝐵 is perpendicular to both 𝑣 and 𝐵 . The magnitude of the magnetic force is simply 𝐹 𝐵 =𝑞𝑣𝐵sin𝜃 Where 𝐹 𝐵 is in newtons, 𝑞 is in coulombs, 𝑣 is in m/s, and 𝐵 is in the unit called tesla (T).

notation The vector 𝐵 is sometimes seen in the perspective shown in the figures. Think of an arrow. Out of the page is like an arrow’s head coming at you. Into the page is like an arrow’s tail going away from you.

Think! Determine the initial deflection of the charged particle. B_in: Up. B_up: Out of the page. B_right: No deflection B_45: Into the page.

Example 1 An electron is accelerated through ∆𝑉=2400 V from rest and then enters a uniform 𝐵=1.70 T magnetic field. What is a.) the maximum value of the magnetic force this particle can experience. b.) the minimum value of the magnetic force this particle can experience. This will be done on the board.

Charged particle in uniform 𝐵-field A positively charged particle is injected into a magnetic field 𝐵 with a velocity 𝑣 exactly perpendicular to the field. The particle is put into circular motion. What’s its radius of motion? Let’s apply Newton’s second law.

Charged particle in uniform 𝐵-field 𝐹 =𝑚𝑎 ⇒ 𝐹 𝐵 =𝑚𝑎 𝑞𝑣𝐵sin𝜃=𝑚𝑎=𝑚 𝑣 2 𝑟 ⇒ Centripetal acceleration Since 𝑣 ⊥ 𝐵 , the angle 𝜃=90°, which gives sin 90° =1. 𝑞𝑣𝐵=𝑚 𝑣 2 𝑟 ⇒ 𝑟= 𝑚𝑣 𝑞𝐵 ⇒

Example 2 A proton (𝑞=1.60× 10 −19 C) is moving in a circular orbit of radius 14 cm in a uniform 0.35 Tesla magnetic field perpendicular to the velocity of the proton. Find the speed of the proton. This will be done on the board.

Example 3 An electron in a television picture tube moves toward the front of the tube with a speed of 8.0× 10 6 m/s along the 𝑥-axis. Surrounding the neck of the tube are coils of wire that create a magnetic field of magnitude 0.025 T, directed at an angle of θ=60° to the 𝑥-axis and lying in the 𝑥𝑦 plane. Calculate the magnetic force on the electron. This will be done on the board.

Lorentz Force 𝐹 =𝑞 𝐸 +𝑞 𝑣 × 𝐵 If a charged particle is in the presence of an electric field 𝐸 and a magnetic field 𝐵 , then the total force (Lorentz force) is 𝐹 =𝑞 𝐸 +𝑞 𝑣 × 𝐵 Let’s take a look at an unintuitive trajectory created by the combination of these forces.

Cycloid motion Suppose that 𝐵 point in the 𝑥-direction and 𝐸 in the 𝑧-direction. A positively charged particle at rest is released from the origin. What path will it follow? The motion of a particle with this motion can be calculated, but it will involve actually using the cross product, which is beyond this course.

Example 4: velocity selector If the speed of a proton is properly chosen, the proton will not be deflected by these crossed electric and magnetic fields. What speed should be selected in this case? This will be done on the board.

mass spectrometer A mass spectrometer separates ions according to their mass-to-charge ratio. The equation for the radius of a charge particle in a 𝐵-field is 𝑟= 𝑚𝑣 𝑞 𝐵 2 And the velocity of an undeflected particle in a velocity selector is Therefore, we can determine m/q by measuring the radius of curvature and knowing the field magnitudes B1, B2, and E. 𝑣= 𝐸 𝐵 1 Therefore the mass-to-charge ratio here is 𝑚 𝑞 = 𝑟 𝐵 2 𝐵 1 𝐸

𝐵-field and current

𝐵-field and current 𝐹 𝐵 = 𝐼 𝐿× 𝐵 𝐹 𝐵 =𝐼𝐿𝐵sin𝜃 The equation for the magnetic force created by a magnetic field 𝐵 on a current-carrying wire with current 𝐼 and length 𝐿 is 𝐹 𝐵 = 𝐼 𝐿× 𝐵 The magnitude of the magnetic force is therefore 𝐹 𝐵 =𝐼𝐿𝐵sin𝜃

Torque on current loop

Torque on current loop 𝜏= 𝐹 2 𝑏 2 sin𝜃+ 𝐹 4 𝑏 2 sin𝜃

Example 5 A rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails that are 𝑑=12.0 cm apart and 𝐿=45.0 cm long. The rod carries a current of 𝐼=48.0 A in the direction shown and rolls along the rails without slipping. A uniform magnetic field of magnitude 0.240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails?

Torque on current loop 𝜏 =𝐼 𝐴 × 𝐵 𝜏=𝐼𝐴𝐵sin𝜃 The general equation for the torque on a current loop placed in a uniform magnetic field 𝐵 is 𝜏 =𝐼 𝐴 × 𝐵 where the area vector 𝐴 is perpendicular to the plane of the loop. Use the Right-Hand Rule to determine the direction of 𝐴 . The magnitude of this equation is 𝜏=𝐼𝐴𝐵sin𝜃

Magnetic dipole moment The magnetic dipole moment 𝑚 is defined, in general, by the following relation: 𝜇 =𝐼 𝑑 𝐴 For a flat current loop, 𝐴 is the area enclosed by the loop, so 𝜇 =𝐼 𝐴 Magnetic dipole moment of a current loop.

Magnetic dipole moment If a coil of wire contains 𝑁 loops of equal area, the magnetic dipole moment becomes 𝜇 =𝑁𝐼 𝐴 The torque on a current-carrying wire loop in magnetic field 𝐵 becomes 𝜏 = 𝜇 × 𝐵

Think! Rank the magnitudes of the torques acting on the rectangular loops (a), (b), and (c) shown edge-on in the figure from highest to lowest. All loops are identical and carry the same current. Rank the magnitudes of the net forces acting on the rectangular loops from highest to lowest.

Example 6 A rectangular coil of dimensions 5.40 cm×8.50 cm consists of 25 turns of wire and carries a current of 15.0 mA. A 0.350 T magnetic field is applied parallel to the plane of the coil. What is the torque on the coil? This will be done on the board.

Example 7 Find the magnetic dipole moment 𝜇 of a “bookend-shaped” loop shown in the figure. The height and width are 𝑑, and the length is 𝑙. The entire loop carries a current 𝐼. This will be done on the board.