physics current

1. Lecture 14 agenda:
 Magnetic Field
 Crossed Fields
 Hall Effect
 Magnetic Dipole Moment

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Magnetic Fields
Sources of Magnetic Field
 Current Carrying conductor
 Permanent Magnet
 Electromagnets
Magnetic Force
The magnetic force acting on a charged particle q, moving with velocity v is given as
𝑭𝑩 = |𝒒|(𝒗 × 𝑩)
𝑭𝑩 = |𝒒|(𝒗𝑩 𝐬𝐢𝐧 𝝋)
Where 𝝋 is angle between directions of 𝒗 𝒂𝒏𝒅 𝑩.
Direction of 𝑭𝑩 is always perpendicular to direction of 𝒗.
If particle is stationary or magnitude of charge is zero, 𝑭𝑩=0.
If direction of 𝒗 𝒂𝒏𝒅 𝑩 is either parallel or anti parallel, 𝑭𝑩=0.
If direction of 𝒗 𝒂𝒏𝒅 𝑩 is perpendicular, 𝑭𝑩 is maximum.

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Magnetic Fields
Direction of Magnetic Force
According to right hand rule
Point your thumb in the direction of velocity, index finger in the direction of magnetic field,
middle finger will point in the direction of magnetic force.
 Cross sign indicates magnetic field is directed into the plane (inward direction).
 Dot sign indicates magnetic field is directed out of the plane (outward direction).
Cross Product of Unit Vectors
𝒊 × 𝒋 = 𝒌, 𝒋 × 𝒌 = 𝒊, 𝒌 × 𝒊 = 𝒋
j× 𝒊 = −𝒌, 𝒌 × 𝒋 = −𝒊, 𝒊 × 𝒌 = −𝒋
Unit
unit of magnetic field is Tesla (T).
1 Tesla=𝟏𝟎𝟒𝑮𝒂𝒖𝒔𝒔

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Crossed Fields
Discovery of Electron

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Crossed Fields
Discovery of Electron
 Charged particles are emitted by a hot filament and accelerated by applied potential difference V.
 After passing through screen C, they form a narrow beam.
 Then they pass through crossed electric and magnetic fields.

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Crossed Fields
Steps
 Set E=0 and B=0, and note the position of the spot on screen due to undeflected beam.
 Turn on E and measure the resulting beam deflection.
 Maintaining E, turn on B and adjust its value in such a way that beam returns to its undeflected
position.
The deflection of particles is given by
𝒚 =
|𝒒|𝑬𝑳𝟐
𝟐𝒎𝒗𝟐

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Hall Effect

8. + -
Hall Effect
Explanation
Consider a copper strip of width d, carrying current i from top to bottom. As charge carriers are
electrons so direction of drift speed 𝒗𝒅 is from bottom to top. Magnetic field is directed into the plane. As
a result, magnetic force acts on electrons, pushing it to the right of strip. As the electrons move from left
to right, leaving positive charges at the left of the strip. This produces an electric field from left to right.
This electric field produces electric force from right to left. Soon after, an equilibrium is achieved. As a
result, force due to E and force due to B are balanced.

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Magnetic Dipole Moment

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Numerical Problems
Exercise Problem 5
An electron moves through a uniform magnetic field given by 𝑩 = 𝑩𝒙𝒊 + 𝟑𝑩𝒙 𝒋. At particular instant,
electron has velocity v= 𝟐𝒊 + 𝟒𝒋 𝒎/𝒔 and magnetic force acting on it is 𝑭𝑩= 𝟔. 𝟒 × 𝟏𝟎−𝟏𝟗
𝒌. Find 𝑩𝒙.
Exercise Problem 9
Consider the following figure. An electron is accelerated from rest through potential difference 𝑽𝟏= 𝟏𝒌𝑽
enters the gap between two parallel plates having separation d=20mm and potential difference 𝑽𝟐=
𝟏𝟎𝟎𝑽. The lower plate is at lower potential. Neglecting fringing effects and assume electron’s velocity
vector is perpendicular to electric field between the plates. In unit vector notation, what uniform
magnetic field allows electron to travel in straight line in the gap?

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Numerical Problems
Exercise Problem 14
A metal strip 6.50cm long, 0.850cm wide, and 0.76mm thick moves with constant velocity v through a
uniform magnetic field B=1.20T directed perpendicular to strip, as shown in figure. A potential
difference of 𝟑. 𝟗𝟎𝝁𝑽 is measured between points x and y across the strip. Calculate speed v.

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Numerical Problems
Exercise Problem 55
Two concentric circular wire loops of radii 𝒓𝟏 = 𝟐𝟎𝒄𝒎 and 𝒓𝟐 = 𝟑𝟎𝒄𝒎 are located in xy plane, each
carries a clockwise current of 7A. a) Find magnitude of net magnetic dipole moment of the system. b)
Repeat for reversed current in inner loop.