Electromagnetism

Magnetic field

Force on a
current-carrying wire

Current on
a balance

Force and torque
on a coil

The galvanometer

Barlow's Wheel

Straight current

The coil

The solenoid and the
toroid

Oscillations of 
a magnet (I)

Oscillations of
a magnet (II)

Magnetic force on a segment of a straight current-carrying wire

Activities

Electric current

Electric current is the charge that passes through the normal section S of a conductor per unit time. In the study of the ionic engine we saw the meaning of mass flux and charge flux or electric current

Let n be the number of particles per unit volume, v the average speed of these particles, S the cross section of the beam and q the charge of each particle.

Then the charge Q that passes through the normal section S in a time t, is that contained in a cylinder of cross section S and length v·t.

Charge Q= (number of particules per unit volume n)·(charge of each particle q)· (volume of the cilinder Svt)

Dividing Q by the time t we obtain the electric current.

The electric current is the flux of charge or the charge that passes through the normal section S per unit time, which is the product of the following terms:

• Number of particles per unit volume, n
• The charge of each particle, q.
• The area of the normal section, S
• The average speed of the particles, v.

Force on a segment of a straight current-carrying wire.

In the mass spectrometer and the cyclotron, we studied the force that exerts a magnetic field on a charged particle and the movement it produces.

The figure shows the direction of the force that exerts the magnetic field B on a positive charged particle q, which moves towards the left with velocity v.

Let us calculate the force on all the charged (nSL) particles contained in the current-carrying wire of length L.

The unit vector u_t=v/v has the same direction as the velocity vector, or the direction in which the positive charged particles move.

In the case of a non-straight current-carrying wire, or a non-constant magnetic field, one has to calculate the force on an element of current dl

• The components of this force dF_x and dF_y
• Check if there is any symmetry that would make one of the components zero
• Finally, the components of the total force F will be calculated by integrating

Activities

To show the force that exerts a magnetic field on an electric current we build a device that consists of a powerful magnet which produces a field of 500 gauss or B=0.05 T, and on its north pole we stick two rails made with copper sheets. The conducting segment is a copper wire of length L=15 cm.

The current is supplied by an arc discharge of i=60 A. As the electric current and the field are perpendicular, the force on the wire will be

F_m=iBL=0.05·60·0.15=0.45 N

If the mass of the wire is 1.35 g, the acceleration will be 333.3 m/s². The large accelerations obtained could suggest the use of the wire as the projectile of an electromagnetic canon.

The velocity of the wire at the end of the 50 cm-long rails is

You can vary

• The intensity of the magnetic field (gauss), in the edit box labelled Magnetic field
• The electric current (A), in the edit box labelled Electric current
• The length of the wire, between 5 and 20 cm, in the edit box labelled Length
• The mass of the wire (g), in the edit box labelled Mass

Press the Start button ,

Observe the movement of the wire. At the top of the applet, you can see the position and velocity of the wire as a function of time.

By pressing the Pause button, you can stop the movement and see the field vector B, the direction of the current, and the force vector F on the wire. To resume the movement of the wire, press the same button which is now labelled Resume.