Motion in Combined Electric and Magnetic Fields: Particle accelerators are one of the most versatile instruments for scientists, especially physicists. From its inception as the cathode ray tube by J.J. Thomson, who used it to discover the electron, to today’s giant colliders, it has been inextricably linked to major milestones in nuclear and particle physics.

Particle accelerators are now widely used in a wide range of applications, including scientific research, applied physics, medicine, and industrial processing, with potential applications in power engineering. This article will discuss everything about Motion in Combined Electric and Magnetic Fields. Scroll down to find more.

Force Acting On A Charged Particle Moving In A Uniform Magnetic Field

The force acting on a particle in a uniform magnetic field of strength \(B\), having a charge \(q\) and velocity \(v\), is given by \({F_B} = q(v \times B)\,{\rm{or}}\,{F_B} = qvB\sin \theta \)

where \({\rm{\theta }}\) is the angle between \(v\) and \(B\)

Case 1: If \(\theta  = 0,{F_B} = 0\). Also if  \(\theta  = {180^\circ },{F_B} = 0\)

If a charged particle enters a uniform magnetic field either in the magnetic field’s direction or opposite to the magnetic field’s direction, the force acting on the charged particle will be zero.

Case 2: If \(\theta  = {90^\circ },{F_B} = qvB\) In this case, the force acting on the particle is greatest, and it acts as a centripetal force, causing the charged particle to move in a circular path.

Therefore, \({F_B} = qvB = {F_{{\rm{centripetal }}}}\)
\(qvB = \frac{{m{v^2}}}{r}\)
\(r = \frac{{mv}}{{qB}}\)

where \(r\) is the circular path’s radius, It is important to note that this force has no effect on the charged particle’s speed or kinetic energy. However, because of the change in direction, the velocity of the charged particle changes, resulting in a change in momentum. In addition, the work done by the force is zero as the force is acting perpendicular to the direction of motion.

Case 3:  If \({\rm{\theta }} = {\rm{is}\,{not}}\,0\,{\rm{or}}\,{90^\circ }\) then the path taken is helical. For better understanding, the velocity of the charged particle can be split into two components:

(a) \(v\cos \alpha \): This is the component which is along the magnetic field, hence force on the charged particle due to this component is zero. This component is responsible for moving the charged particle uniformly in the direction of \(B\).

(b) \(v\sin \alpha \) : This is the component which is perpendicular to the magnetic field,The force acting on charged particle due to this component is given by;
\({F_B} = q(v\sin \alpha )B\sin {90^\circ } = qvB\sin \alpha \)

This acts as the centripetal force, causing the particle to move in a circular path. The result of these two effects is a helical path. When a charged particle enters a uniform magnetic field at an angle, it follows a helical path.

Radius of the helix,

\(r = \frac{{mv\sin \alpha }}{{qB}}\)

Angular frequency of rotation;

\(\omega  = \frac{{2\Pi }}{T} = \frac{{qB}}{m}\)

Pitch of the helix;

\( = v\cos \alpha *T = \frac{{2\Pi mv\cos \alpha }}{{qB}}\)

Motion in Combined Electric and Magnetic Fields

When electric and magnetic fields act upon a moving charge, the charge experiences the Lorentz force, a vector sum of forces due to electric and magnetic fields.

\(F = {F_E} + {F_B}\)
Where \({F_E} = qE\)
and \({F_B} = q(v \times B)\)

Lets us look at the two possible cases:

Case: 1 (When E, B and v all three are colinear)

We can see here as the particle is moving along the magnetic field, the magnetic force on it will be zero, no matter what is the magnitude or nature of the charge. Only force that the charged particel will experience would be an electric force whose direction would depend on the nature of the charge. If the mass of the particle is known, we can find its acceleration as;

\(a = \frac{{{F_E}}}{m} = \frac{{qE}}{m}\)

This clearly suggests that the particle will move in a straight line without any change in its direction, though its velocity will keep on changing, leading to a change in its kinetic energy and momentum.

Case: 2 (When E, B and v are mutually perpendicular)

When the electric field, the magnetic field, and the motion of charge are mutually perpendicular to each other (as shown in the diagram below), then they are called crossed fields, and forces due to electric and magnetic fields will act in the opposite directions, so the Lorentz force F is given by;

\(F = qE\hat i + (qv\hat i \times B\hat k)\)
\(F = qE\hat j – qvB\hat j\)
\(F = q(E – vB)\hat j\)

When the strength of the electric and magnetic fields is varied to equalise the forces due to the electric and magnetic fields, the charge can move in the field without deflection. This concept is used in velocity selectors.

What is the Velocity Selector?

While studying the motion of charged particles in a uniform magnetic field, we know that the charged particles will experience a force due to both electric and magnetic fields. When we consider a charge beam, we know that the velocities of all the charged particles will not be the same. In some experiments, we want a specific charge with a specific velocity; to obtain such charged particles, we use velocity selectors.

The velocity selector, also known as the Wien filter, is a configuration of an electric and magnetic field. The arrangement of the electric and magnetic fields is used to select a charged particle with a specific velocity from a beam of charges moving at different speeds, regardless of their mass or charge.

Consider a charged particle of charge \(q\) moving with velocity v in uniform electric and magnetic fields, where the electric field, magnetic field, and charged particle velocity are all perpendicular to each other. Consider the electric field to be along the \(y\)-axis, the magnetic field to be along the \(z\)-axis, and the charge velocity to be along the \(x\)-axis.

The charged particle considered will experience the force due to both magnetic and electric fields. We know that the force exerted on charge \(q\) is given by
\({F_E} = qE = qE\hat j\)

Similarly, the force exerted by the magnetic fields is given by:
\({F_B} = \;q\left( {v \times B} \right) = q\left( {{v_0}\hat i\; \times B} \right)\hat k = q{v_0}B\;\left( { – \hat j} \right)\)

Clearly, the electric force and magnetic force are opposite in direction. Now, for the velocity selector, we follow the condition where both forces must be equal to each other. Then from equation (1) and (2), we get,
\( \Rightarrow qE\hat j = q{v_0}B( – \hat j)\)
\({v_0} = E/B\)

In a velocity selector, charged particles must move with a speed of \({v_0} = EB\)

in order to pass through the equipment. Hence, as its name suggests, the velocity selector allows charged particles with a particular velocity to pass through (hence, selecting particles of a certain velocity). The mechanism of a velocity selector is shown in the below figure.

A uniform electric and magnetic field will surround the velocity selector. Consider the mechanism depicted in the figure above: a positively charged top plate and a negatively charged bottom plate will generate a uniformly charged electric field.

This will result in the formation of an electric field between the two plates, pointing downward. In addition, a uniform magnetic field will be generated between the plates. The uniform magnetic field can be directed in either direction. To put it another way, the uniform magnetic field can be directed into or out of the paper. The magnetic field shown in the figure above is directed into the paper.

The upwards force must equal the downwards force in order for the charged particle to pass through space without being deflected (either upwards or downwards). If the positively charged particle has a slightly higher velocity than \(E/B\), it will be deflected upwards due to the greater upward force.

There is one more particle accelerator which uses both magnetic and electric fields for it to function, it’s called a cyclotron.

Solved Examples on Motion in Combined Electric and Magnetic Fields

The solved examples on Motion in Combined Electric and Magnetic Fields are given below:

Q.1. A Velocity Selector is used to select \(200\;{\rm{KeV}}\) Alpha Particles from a Beam of Particles of various energies. The strength of the electric field is \(1200\;{\rm{ kV/m}}\). How Strong must the magnetic field be?
Solution:
Given,
Kinetic Energy of the alpha particle \( = 200\;{\rm{KeV}}\)
We know that the mass of an alpha particle is \(6.68 \times {10^{ – 27}}\;{\rm{kg}}\). Thus, the velocity of the alpha particle is given by
\(v = \sqrt {\frac{{2k \cdot E}}{m}} \)
\(v = 3.095 \times {10^6}\;{\rm{m}}/{\rm{s}}\)
We know that for a velocity selector \({v} = E/B\)
Therefore, the magnetic field strength for the particle moving with velocity \(3.095 \times {10^6}\;{\rm{m}}/{\rm{s}}\)
\(B = E/v\)
\(B = \frac{{1200 \times {{10}^3}}}{{3.095 \times {{10}^6}}} = 387\;{\rm{mT}}\)
Where,
\(E\) is the strength of the electric field, and \(B\) is the strength of the magnetic field
Therefore, the magnetic field strength is \(387\;{\rm{ mT}}\).

Summary

Electrical and magnetic fields can be used in tandem to accelerate and control charged particles. Several types of particle accelerators use these fields to achieve the desired path or velocity of the particles. Velocity selectors, as their name suggests, allow only charged particles with a particular velocity to pass through. Charged particles can also execute helical paths inside a magnetic field under certain conditions. The frequency of revolution depends on the magnitude of the charge, the magnetic field, and the mass of the particle.

FAQs on Motion in Combined Electric and Magnetic Fields

The frequently asked questions on Motion in Combined Electric and Magnetic Fields are given below:

Q.1. What is the trajectory of a charged particle moving in the direction of a uniform magnetic field?
Ans: The charged particle will move along a straight line path because the magnetic force experienced by it will be zero.

Q.2. What is a velocity selector?
Ans: The velocity selector, also known as the Wien filter, is used to detect and isolate charged particles moving at a specific speed.

Q.3. What is the path of a charged particle moving inside a uniform magnetic field at having a velocity at a certain angle to the field?
Ans: A charged particle entering a uniform magnetic field at any angle except \(0^\circ ,{\rm{ }}180^\circ \), and \(90^\circ \). executes a helical path.

Q.4. Which force provides the necessary centripetal acceleration when a charged particle executes circular motion when projected in a uniform magnetic field?
Ans: Magnetic force provides the necessary centripetal force for the charged particle to execute the circular motion.

Q.5. What is the use of cyclotron?
Ans: Heavy charged particles such as protons, deuterons, and heavier ions are accelerated in cyclotrons.

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