Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two long wires with currents in the same direction attract each other. Now take two electron beams moving side by side, in the same direction. Each beam is a current, and the two currents are parallel. So the beams attract, right? Hold on to that answer. Today we do seven concept checks from this unit's problem bank. Questions like these come up often in JEE, so let's begin.

0:30Check 1: can a magnetic field change kinetic energy?

So, check one. The force on a charge q, moving with velocity v in a magnetic field B, is F equals q times v cross B, okay? The arrows on top mean vectors, and the cross is the cross product. This force is always perpendicular to both v and the field. Now here is the check. Can a static magnetic field change the kinetic energy of a charged particle? So let us think about work. The upside-down T means perpendicular: the force is always at right angles to the motion, so it does no work at all. Right? No work, so no change in kinetic energy. The speed stays constant, and only the direction changes. So the answer is no. Clear?

1:17Check 2: does a velocity selector care about q or m?

Now check two, the velocity selector. The electric field, E, and the magnetic field, B, are perpendicular to each other and to the velocity, v, set up so that the two forces on the particle are opposite. With me? Here is the check. Does the selected speed depend on the charge, q, or the mass, m? The particle goes straight when the two forces balance: q times E equals q times v times B, okay? Cancel q, and we get v equals E over B, simple! No q and no m in it. So the answer is no: whatever its charge or mass, a particle with this speed goes straight. Got it?

1:59Check 3: a current loop in a uniform field

Now check three. The figure shows a rectangular coil edge-on, in a uniform field B, the blue arrows. The dot in a circle is current out of the page; the cross in a circle, current into it. The top side feels the force F, equal to I times l times B for each turn, upward, with l the length of that side; the bottom side feels an equal force downward. Okay? Here is the check. In a uniform field, does a current loop feel a net force? And a torque? First the force. For a loop of any shape in a uniform field, the forces on its elements add up to zero, like these two equal and opposite forces. So, no net force. Right? But the two forces do not act along one line: they are b sine theta apart, marked at the bottom, b being the side seen edge-on. The green arrow is mu, the Greek letter, the magnetic moment, N times I times A times n hat, the unit normal, for N turns of area A each. Theta is the angle between mu and the field. So the torque, tau, the Greek letter, is mu cross B, of size mu times B times sine theta, as on the right, and we skip the energy U there. With me? So when is the torque zero? Only when sine theta is zero, theta zero or one hundred and eighty degrees: mu parallel to the field, the two upright lines on screen. So, no net force, but a torque, unless the two are parallel. Clear?

3:37Check 4: zero field inside a tube and a toroid

Now check four, using Ampere's law; we will use the figure below in a moment. The integral of B dot d l around a closed loop, the circle on the integral sign, equals mu naught times I enclosed. Here d l is a small step along the loop, mu naught, mu with a small zero, is the permeability of free space, and I enclosed, the I with the small letters e n c, is the net current through the loop. It helps only when symmetry keeps the field the same all along the loop. Okay? Here is the first question. Take a long, thin-walled tube carrying a current along its length. Is the field inside it zero? Yes, it is zero! Draw a circle inside the tube, coaxial with it. All the current flows in the wall, outside this circle, so I enclosed is zero. By symmetry the field is the same size all round, so it must be zero. Got it? Now the second question: the toroid, on the right of the figure; we skip the solenoid on the left. The toroid is a solenoid bent into a ring, the shaded core, with N turns carrying a current I, okay? On the dashed circle of radius r, inside the core, the field is mu naught N I over two pi r. With me? Now the check. What is the field just outside the toroid, and in its central hole? Zero in both, exactly! Draw a circle in the central hole: it encloses no current. Draw one just outside: each turn crosses the surface inside it twice, once going up and once coming down, so the net current is zero again. Clear?

5:18Example 1: does cyclotron frequency depend on speed?

Now, Example one, the easier one. Does the cyclotron frequency depend on the particle's speed? Pause the video and try it first. Done? Okay, let's check. Surely a faster particle goes round more often, correct? Oh no, careful, the circle grows. The magnetic force, q v B, provides the centripetal force, m v squared over r. The double arrow means which gives: the radius r equals m v over q B, so a faster particle moves in a bigger circle. Right? The period, T, the time for one round, is the distance two pi r over the speed v. Put in r, and v cancels: T equals two pi m over q B, okay? So the cyclotron frequency, f, is one over T, which gives q B over two pi m, for speeds much less than c, the speed of light; the double less-than sign means much less than. With me? So the answer is no. Twice the speed gives twice the radius, and the same time for each round. Very good!

6:28Example 2: do parallel electron beams attract or repel?

Now Example two, a little harder, and it is our puzzle from the start. Two parallel electron beams move side by side in the same direction. Do they attract or repel? Pause the video and try it first. Done? Okay, let's check. Both beams are currents in the same direction, and parallel currents attract. So the beams attract, right? That is only half the story. Yes, parallel currents give a magnetic attraction. But that is not the only force here, okay? The beams are made of electrons, and every electron is negative. Like charges repel, so there is also an electric repulsion. Right? So now it is a tug of war. Which force is stronger, the magnetic attraction or the electric repulsion? Look at the ratio: the magnetic attraction over the electric repulsion is v squared over c squared, with v the speed of the electrons and c the speed of light. Since v is less than c, the ratio is less than one, the less-than sign, so the attraction is weaker. Clear? So the electric repulsion wins, and the beams repel. Each beam is a current, but it is also a line of charge, and that solves our puzzle!

7:45Quick revision: no work, v = E/B, loop torque, cyclotron

So, quick revision. The magnetic force does no work, and a selector picks v equals E over B, for any charge and mass. In a uniform field, a loop feels no net force but a torque, mu cross B, and with symmetry and no current enclosed, the field is zero. The cyclotron frequency, q B over two pi m, does not depend on speed, and parallel electron beams repel. Now try the practice problems on this page. All the best!