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Notes

• Homework # 6 has been posted on the course webpage

• This will be the last homework before mid-sem

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Summary of Lecture # 16:

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• Magnetic Force:  mag = Q( × )  Lorentz Force Law•

The work done by a magnetic force is zero !

mag = � ×  

= =   mag =  × Q = � ×  

mag =  × Q = � ×  

= =  

=   =  =

  � ×

 

• Current

• Surface

Current

Density

• Volume

Current

Density

•

The Continuity Equation ⋅ =   

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The Biot-Savart Law

The magnetic field produced by a steady line current

= 04 �′×  

 ̂r 2  •   0 is the permeability of free space•   0 = 4 × 10−7 N/A2 • The unit of magnetic field is Newton per Ampere-meter, or Tesla

• 1 Tesla is a very strong magnetic field. Earth’s magnetic field is

about 10−4 times smaller

• Biot-Savart law for magnetic field is analogous to Coulomb’s law for

electric field

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The Biot-Savart Law

The magnetic field produced by a surface current

= 04�(

) ×  

 ̂r 2   ′ 

The magnetic field produced by a volume current

= 04

� () ×   ̂r 

2  ′ 

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The Biot-Savart Law

=   

= cos2  

×  

 ̂=

 cos

 

=  tan 

=   r  cos  1r 2 =

cos2

2  

= 04   �

cos2 2

cos2 cos 

  = 0

4   � cos 

− 

=04

  sin

2+ sin

1  

Ex. 5.5 (Griffiths, 3rd   Ed. ): Calculate the

magnetic field due to a long straight wire

carrying a steady current

 .

= 04 �′

×   ̂r 2  

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The Biot-Savart Law

Ex. 5.5 (Griffiths, 3rd   Ed. ): Calculate the

magnetic field due to a long straight wire

carrying a steady current

 .

= 04 �

′ ×   ̂r 2  

=04

  sin2 + sin1  

Field due to an infinite wire ?

2 = 2 1 =2

 

  =

04

  sin

2+ sin

1  

=04   1 + 1  

=

02   7

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The Biot-Savart Law

Ex. 5.6 (Griffiths, 3rd   Ed. ): Find the   fieldabove the center of a loop (radius , current ).

= 04 �′×  

 ̂r 2  The line element ′ produces the field  at . Thehorizontal components of this field cancels out.

= 04  ′r 2 cos 

The total field, which is in the  direction, is

= 04 �cos

r 2   ′  = 04cos

r 2   �′  =04

r 3 2  

=02

22 + 2 3/2 

What is the field at the center?

0 =

02

2

3  =

02

 

The vertical component of this field is

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The Curl of  What is the curl of  ? Should be non-zero !

Check for the case of straight wire with current

 

 

⋅  

B  = 02 

=

  02

 =02

 

For circular path of radius  =

02

2

  =

 0 

The line integral is independent of  For an arbitrary path enclosing the current carrying wire

The field is best represented in the cylindrical coordinate

= 02   =   +  

+   

 

⋅   =

  02 ⋅

(

 

+

 

+

 

)  =

02 �

2

0  =

 0 So,

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The Curl of  What is the curl of  ? Should be non-zero !

  ⋅   = 0 Check for the case of straight wire with current  

If the path encloses more than one current carrying wire

  ⋅   = 01 + 02 + 03 = 0enc enc = � ⋅  But

 

⋅ =

 0 � ⋅ 

�( × ) ⋅ = 0 � ⋅   × = 0 • Is this valid only for straight wires?  No

• It is valid in general 10

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The Ampere’s Law

× = 0 

  ⋅ = 0enc 

Ampere’s law in differential form

Ampere’s law in integral form

• Ampere’s law is analogous to Gauss’s law

• Ampere’s law makes the calculation of magnetic

field very easy if there is symmetry.

• If there is no symmetry, one has to use Biot-Savart

law to calculate the magnetic field.

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The Ampere’s Law

Ex. 5.5 (Griffiths, 3rd   Ed. ): Calculate the

magnetic field due to an infinitely long

straight wire carrying a steady current

 .

=02    

We want to do this problem using Ampere’s law

Using Biot-Savart Law

Make an Amperian loop of radius  enclosing the current

Since it is an infinite wire, the magnetic field must

 be circularly symmetric. Therefore,

  ⋅   = 2  = 0 =   = 0enc 

= 02  So Easy !! 12