5. Magnetic Field1 By Sanjay Pandey

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5. Magnetic Field

1.

Magnetic Field

β€’

One of the biggest differences is that electrical charges can be isolated from each other (a negative charge can be sitting all alone), while magnetic poles must come in pairs (north and south) Did You Know? There are some theories in modern physics that indicate that it should be possible (even though never been done) to isolate a north pole from a south pole. The dipoles would become

We can imagine a magnetic field surrounding a magnet in much the same way that we did for electrical charges. β€’

2.

monopoles.

Similarities and differences Magnetic Fields Strong field

Gravitational Fields

Weakest of all fields

Electric Fields Strong field.

Not directly calculated in Physics (although we do measure it indirectly)

Calculated using an inverse square law (Newton's Universal Law of Gravitation)

Calculated using an inverse square law (Coulomb's Law)

Directly related to the magnet involved

Directly related to the masses involved

Directly related to the charges involved

Follows inverse square law near the magnet but follows an inverse cubed law further away so that the field becomes exponentially weaker as separation increases

Follows inverse square law so that the field becomes exponentially weaker as separation increases

Attraction or Repulsion

Individual poles can never be separate from each other

Always attraction

Individual masses are separate from each other

Attraction Repulsion.

or

Individual charges are separate from each other Follows inverse square law so that the field becomes exponentially weaker as separation increases.

β€’ Magnetic field exerts force on a charge particle β€’ Some facts about the magnetic force a) From a point P, a charged particle can move in any direction or along any line. Along one of these possible lines, if the charge is moving, there is no magnetic force. Magnetic force is defined to be acting along this line. b) The magnitude of the magnetic force is proportional to the product of speed of the charged particle 𝑣𝑣 and 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠, πœƒπœƒ being the angle the speed makes with the line along which magnetic field is acting. Hence magnetic force is proportional to |𝑣𝑣 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠|. c) The direction of the magnetic force is perpendicular to the direction of the magnetic field as well as to direction of the velocity. d) The magnetic force is also proportional to the magnitude of charge π‘žπ‘ž. e) Its direction is different and opposite for positive and negative charges. Magnetic force can be defined mathematically as

οΏ½βƒ— 𝐹𝐹⃗ = π‘žπ‘žπ‘£π‘£βƒ— Γ— 𝐡𝐡

β€’

οΏ½βƒ— from the rules of the vector product. Equation uniquely determines the direction of magnetic field 𝐡𝐡

Characteristics of the Force

2 A magnetic field can create a force on an object. However, for the object to feel a force, and the

magnetic field to affect the object, three things must be true 1.

The object must have an electric charge.

2.

The charged object must be moving.

β€’

direction of the magnetic field. Units of magnetic field

3.

The velocity of the moving charged object must have a component that is perpendicular to the The SI unit of magnetic field is 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛/π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž βˆ’ π‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘š. It is written as 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇. 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 is 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛/π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž βˆ’ π‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘š. Tesla is also defined as 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀/π‘šπ‘šΒ². Another unit in common use is 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 .

β€’ β€’ β€’ 3.

1 𝑇𝑇 = 104 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

We have magnetic field of the order of 10βˆ’5 near the earth's surface. Superconducting magnets can create a magnetic field of the order of 10 𝑇𝑇. Earlier, the concept of magnetic field was referred to as magnetic induction.

Electromagnetic field Electric field and magnetic field are not basically independent. They are two aspects of same entity electromagnetic field. Whether the electromagnetic field will show up as an electric field or a magnetic field or a combination depends on the frame from which we are looking at the field.

Note: We represent magnetic field vectors like that as arrows. But all we see is either the tip of the arrow βŠ™, if the field is coming out of the page, or the tail of the arrow, βŠ—, if the field is going into the page. 4.

Motion of a Charged particle in a uniform magnetic field Magnetic force on a charged particle is perpendicular to its velocity. Hence there will not any change in its speed or kinetic energy.

The magnetic force will deflect the particle without changing speed and in a uniform field, the particle will move along a circle perpendicular to the magnetic field. The conclusion is that, the magnetic force provides centripetal force. If r be the radius of the circle, then π‘žπ‘žπ‘žπ‘žπ‘žπ‘ž = π‘šπ‘šπ‘šπ‘šΒ²/π‘Ÿπ‘Ÿ

3 (LHS is the expression for magnetic force and RHS is expression mass Γ— acceleration) π‘Ÿπ‘Ÿ = π‘šπ‘šπ‘šπ‘š/π‘žπ‘žπ‘žπ‘ž

The time taken to complete the circle is 𝑇𝑇 = 2πœ‹πœ‹πœ‹πœ‹/𝑣𝑣 = 2πœ‹πœ‹πœ‹πœ‹/π‘žπ‘žπ‘žπ‘ž

The time period or time taken to complete one circle is independent of speed. But the radius depends on 𝑣𝑣. Hence if speed increases, the radius is larger.

Frequency of revolutions is

𝜈𝜈 = 1/𝑇𝑇 = π‘žπ‘žπ‘žπ‘ž/2πœ‹πœ‹πœ‹πœ‹

This frequency is called cyclotron frequency. β€’

Helical Paths If the velocity of charge is not perpendicular to the magnetic field, the resultant path will be a helix.

The radius of the path will be determined by velocity component which is perpendicular to the magnetic field. If πœ™πœ™ is the angle between 𝑣𝑣 and 𝐡𝐡, then there are two components of velocities

(i) (ii)

𝑣𝑣 sin πœ™πœ™ perpendicular to magnetic field B this component provides circular motion about B 𝑣𝑣 cos πœ™πœ™ parallel to the magnetic field B this component provides motion of translation

The radius of helix

π‘šπ‘š (𝑣𝑣 sin πœ™πœ™ )2 π‘Ÿπ‘Ÿ

= π‘žπ‘žπ‘žπ‘žπ‘žπ‘ž sin πœ™πœ™

4 or

π‘Ÿπ‘Ÿ =

π‘šπ‘šπ‘šπ‘š sin πœ™πœ™ π‘žπ‘žπ‘žπ‘ž

The time taken to complete one revolution is 𝑇𝑇 =

5.

2πœ‹πœ‹πœ‹πœ‹

𝑣𝑣 sin πœ™πœ™

=

2πœ‹πœ‹πœ‹πœ‹ π‘žπ‘žπ‘žπ‘ž

T is independent of π‘Ÿπ‘Ÿ, 𝑣𝑣 π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž 𝛼𝛼. 2πœ‹πœ‹πœ‹πœ‹ ∴ Pitch 𝑝𝑝 = 𝑣𝑣 cos πœ™πœ™ . 𝑇𝑇 = 𝑣𝑣 cos πœ™πœ™ .

or

Pitch

𝑝𝑝 = 𝑣𝑣 cos πœ™πœ™ .

2πœ‹πœ‹πœ‹πœ‹

𝑣𝑣 sin πœ™πœ™

π‘žπ‘žπ‘žπ‘ž

= 2πœ‹πœ‹πœ‹πœ‹ cot πœ™πœ™

Magnetic Force on a current carrying wire

In a current carrying wire, electrons, which are charge carrying particles are moving and hence in a magnetic field, a current carrying conductor would experience magnetic force. on it is 6.

If a straight wire of length 𝑙𝑙 carrying a current 𝑖𝑖 is placed in a uniform magnetic field B, then the force οΏ½βƒ— 𝐹𝐹⃗ = 𝑖𝑖𝑙𝑙⃗ Γ— 𝐡𝐡

The quantity 𝑖𝑖𝑖𝑖 denotes current element of length of 𝑙𝑙.

Torque on a current loop

If there is a rectangular loop carrying current 𝑖𝑖 in a uniform magnetic field B then net torque acting on the loop is Π“ = 𝑖𝑖𝑖𝑖𝑖𝑖 sin πœƒπœƒ

Where, 𝑖𝑖 = current in the loop 𝐴𝐴 = area

B = magnetic field

πœƒπœƒ = the angle of inclination of the loop with the plane perpendicular to the plane of magnetic field. We can also define

οΏ½Π“βƒ— = 𝑖𝑖𝐴𝐴⃗ Γ— 𝐡𝐡 οΏ½βƒ—

𝑖𝑖𝑖𝑖 can be termed as πœ‡πœ‡ the magnetic dipole moment or simply magnetic moment of the current loop. If there are 𝑛𝑛 turns in the loop, each turn experiences a torque. The net torque is

7.

field.

οΏ½Π“βƒ— = 𝑛𝑛𝑛𝑛𝐴𝐴⃗ Γ— 𝐡𝐡 οΏ½βƒ— πœ‡πœ‡βƒ— = 𝑛𝑛𝑛𝑛𝐴𝐴⃗ Lorentz force: οΏ½βƒ— and electric field 𝐸𝐸�⃗ experiences a force A moving charge in presence of a magnetic field 𝐡𝐡 βƒ— βƒ— βƒ— οΏ½βƒ— οΏ½βƒ— 𝐹𝐹 = 𝐹𝐹𝑒𝑒 + πΉπΉπ‘šπ‘š = π‘žπ‘žπΈπΈ + π‘žπ‘žπ‘£π‘£βƒ— Γ— 𝐡𝐡 οΏ½βƒ—) β‡’ 𝐹𝐹⃗ = π‘žπ‘ž(𝐸𝐸�⃗ + 𝑣𝑣⃗ Γ— 𝐡𝐡 οΏ½βƒ— and 𝐹𝐹⃗𝑒𝑒 = βˆ’πΉπΉβƒ—π‘šπ‘š and resultant field is called crossed If Lorentz force is zero then 𝐸𝐸�⃗ should be βŠ₯ to 𝐡𝐡

Case I: οΏ½βƒ— all the three are collinear: When 𝑣𝑣⃗, 𝐸𝐸�⃗ and 𝐡𝐡 In this situation as the particle is moving parallel or anti parallel to the field (i.e., πœƒπœƒ = 0 Ν¦ or 180 Ν¦), the magnetic force on it will be zero and only electric force will act and so 𝐹𝐹 π‘žπ‘žπ‘žπ‘ž π‘Žπ‘Ž = = π‘šπ‘š

π‘šπ‘š

Hence the particle will pass through the field following a straight line path with change in speed.

5 Case II:

οΏ½βƒ— are mutually perpendicular: When 𝑣𝑣⃗, 𝐸𝐸�⃗ and 𝐡𝐡 οΏ½βƒ— are such that In this situation if 𝐸𝐸�⃗ and 𝐡𝐡 𝐹𝐹⃗ = 𝐹𝐹⃗𝑒𝑒 + πΉπΉβƒ—π‘šπ‘š = 0 i.e., π‘Žπ‘Ž =

𝐹𝐹

π‘šπ‘š

= 0 or cross field

οΏ½βƒ—. The particle will move undeflected perpendicular to 𝐸𝐸�⃗ and 𝐡𝐡

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