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Handout 7: Magnetic force
Magnetic force on moving charge
A particle of charge 𝑞 moving at velocity 𝐯 in magnetic field 𝐁 experiences a magnetic force
𝐅 = 𝑞𝐯 × 𝐁.
The direction of the magnetic force is the same as the direction of the cross product 𝐯 × 𝐁 . This is shown in
Fig. 1. The direction of 𝐅 is always perpendicular to both 𝐯 and 𝐁 . Since magnetic force is perpendicular to velocity, work done by the force is zero. Magnetic force never does work on the particle.
The magnitude of the magnetic force is given by
𝐹 = 𝑞𝑣𝐵 sin 𝜃, where 𝜃 is the angle between vector 𝐯 and vector 𝐁 . If the particle moves along with the magnetic field, 𝜃 = 0° or 180° , the magnetic force vanishes.
If the particle moves perpendicular to the magnetic field, 𝜃 = 90° and the magnitude of the magnetic force becomes 𝐹 = 𝑞𝑣𝐵 . The direction of the force is perpendicular to velocity. As a result, the particle undergoes a circular motion . The magnetic force 𝐹 = 𝑞𝑣𝐵 provides the centripetal force. Therefore, 𝑞𝑣𝐵 = 𝑚𝑣 2
𝑅
⟹ 𝑅 = 𝑚𝑣 𝑞𝐵
.
Example 1 A particle of charge 𝑞 and mass 𝑚 performs a circular motion in a uniform magnetic field 𝐵 . Show that the period of the motion is given by
𝑇 =
2𝜋𝑚 𝑞𝐵
.
Example 2 A particle undergoes uniform circular motion of radius 28.7 𝜇 m in a uniform magnetic field. The magnetic force on the particle has a magnitude of
1.6 × 10 −17 N. What is the kinetic energy of the particle?
Figure 1: Direction of the cross product
Figure 2: Circular motion of charged particle

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Example 3 A particle of charge +𝑒 and mass 𝑚 is ejected from the origin at velocity 𝑣 at angle 𝜃 with the 𝑥 -axis. A uniform magnetic field 𝐵 is also along the 𝑥 -axis. The particle performs a helical motion. Find the “pitch” P of the helix.
Example 4 Two isotopes of helium nuclei, 3 He and 4 He, are accelerated through the same potential difference.
They enter the region of uniform magnetic field at S
2
and follow semicircular paths with different radii. Determine the ratio of the radius of 3 He path to the that of 4 He path.
Example 5 The diagram shows the Thomson’s apparatus for measuring the charge-to-mass ratio of an electron.
An electron (charge 𝑒 , mass 𝑚 ) is accelerated from cathode (C) to anode (A) through a voltage 𝑉 . Then it enters the region with perpendicular 𝐸 -field and 𝐵 -field chosen such that the electron moves along a straight light without deflection and hits the screen.
Show that 𝑒 𝑚
=
𝐸 2
2𝑉𝐵 2
.

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Hall effect
Consider a conductor in the form of a flat strip, as shown in Fig. 3. The current density 𝐽 𝑥
is in the direction of the +𝑥 -axis and there is a uniform magnetic field 𝐁 perpendicular to the plane of the strip, in the +𝑦 direction. The drift velocity of the moving charges has magnitude 𝑣 𝑑
. Fig. 3(a) shows the case of negative charges such as electrons in metals and Fig. 3(b) shows the case of positive charges. In both cases, the magnetic force is upward, driving the charge to the upper edge.
The direction of the electric field depends on the sign of the charge carriers.
If the charge carriers are electrons [Fig. 3(a)], an excess negative charge accumulates at the upper edge, leaving the lower edge positive. The accumulation continues until the resulting transverse electric field 𝐄 is large enough to produce electric force 𝑞𝐸 to balance with the magnetic force 𝑞𝑣 𝑑
𝐵 𝑦
. After that, there is no longer deflection of the moving charge. The produced electric field causes potential difference between the opposite edges of the strip, called “Hall voltage” or
“Hall emf” .
Consider Fig. 3(b), when electric force and the magnetic force are balanced. From 𝐅 = 𝟎 , 𝑞𝐸 𝑧
+ 𝑞𝑣 𝑑
𝐵 𝑦
= 0 ⟹ 𝐸 𝑧
= −𝑣 𝑑
𝐵 𝑦
.
We can eliminate 𝑣 𝑑
by using 𝑣 𝑑
= 𝐽 𝑥 𝑛𝑞 number density of the carriers. Therefore,
where 𝑛 is the 𝑛𝑞 =
−𝐽 𝑥
𝐵 𝑦
𝐸 𝑧
.
This confirms that when 𝑞 is positive, 𝐸 𝑧
Similarly, when 𝑞 is negative, 𝐸 𝑧
is negative.
is positive. The sign of the charges is determined by the polarity of the Hall voltage.
Figure 3: Hall effect with (a) negative charge carriers and (b) positive charge carriers

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Example 6 A metal strip with 𝑤 = 11.8
mm and 𝑡 = 0.23
mm, carries a current of 78 A. The strip lies in a uniform magnetic field 𝐵 = 2.29
T perpendicular to the plane of the strip. The Hall voltage is found to be 130 𝜇 V. Find the density of free electrons in this metal.
Magnetic force on current-carrying conductor
Consider an electric current 𝐼 flowing in a straight conductor as shown in Fig. 4. The positive charge is moving with drift velocity 𝐯 𝑑
. The magnetic force on the positive is given by 𝐅 = 𝑞𝐯 𝑑
× 𝐁 to the left.
We can find the total force on the conductor by replacing 𝑞 by the total charge 𝑄 = 𝐼𝑡 within time 𝑡 . Hence, the total force can be expressed as 𝐅 = 𝐼 𝑡𝐯 𝑑
× 𝐁 , yielding
𝐅 = 𝐼𝓵 × 𝐁, where vector 𝓵 = 𝑡𝐯 𝑑
represents the length of the wire in the direction of the current.
The magnitude of the magnetic force is given by
𝐹 = 𝐼ℓ𝐵 sin 𝜃, where 𝜃 is the angle between vector 𝓵 and vector 𝐁 . If the conducting wire is aligned with the magnetic field, 𝜃 = 0° or 180° , the magnetic force is zero. If the wire is perpendicular to the magnetic field, 𝜃 = 90° and the magnitude of the magnetic force becomes 𝐹 = 𝐼ℓ𝐵 .
Example 7 A wire carries a current of 10.0 A in a direction that makes an angle of 30.0° with the direction of a magnetic field of strength 0.300 T. Find the magnetic force on a 5-m length of the wire.
Figure 4: Force on a moving positive charge in a current-carrying conductor

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Example 8 A conductor suspended by two flexible wires as shown in the figure has a mass per unit length of
0.040 0 kg/m. What current must exist in the conductor for the tension in the supporting wires to be zero when the magnetic field is 3.60 T into the page? What is the required direction for the current?
Example 9 A rectangular loop ABCD is placed in a uniform magnetic field 𝐵 parallel to the plane of the loop. The length of sides AB and CD is equal to a and the length of sides BC and DA is b . A current 𝐼 passes through the loop. Find the torque due to magnetic forces about the axis.
Example 10 A conducting cylinder of length 𝐿 has total charge 𝑄 uniformly distributed on the curved surface.
The cylinder is rotating about its symmetry axis with angular speed 𝜔 and there is a uniform magnetic field 𝐵 parallel to the cylinder axis. Show that the pressure on the curved surface is given by 𝑝 =
𝑄𝜔𝐵
2𝜋𝐿
.