Ch-29-Magnetic Fields Due to Currents

8/13/2019 Ch-29-Magnetic Fields Due to Currents

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Biot-Savart law:

dB = ( mo/4p) i d sx /r 2

mo = 4 p 10 -7 T m/A

29.2: Calculating the magnetic field due to a current:

BSL is an inverse square law!!

If d s and are parallel, the contribution is zero!


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Special cases:

For a straight wire making angles q1 and q2:

B = ( mo i/4pa)(cos q1- cos q2)

grasp the element in your right hand with your extended thumb pointingin the direction of the current. Your fingers will then naturally curl

around in the direction of the magnetic field lines due to that element.

The direction is found through the right hand rule:

Prove it!


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For infinite straight wire:

B = ( mo i/2pa)

At the center of a circular arc with angle f :

The radial part extending from the arc does not contribute; why?

At the center of a circular loop:

B = ( mo i/2R)

Checkpoint #1

B = ( mo i f /4pR)


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Ba = moia/(2pd)

Therefore, the force felt at wire #b is:

29.3: Force between two parallel currents:

For two long, straight parallel wires a distance [d] apart, themagnetic field created by current ia at wire #b is:

F ba = I b L x Ba

L and B are perpendicular, therefore F ba

= i b

L Ba

F ba/L = moiai b/(2pd)


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What about the force which wire #a feels?

The same!! Why?

Fab/L = moiai b/(2pd)

In which direction are the forces?

Towards one another if the currents are parallel and away from oneanother if they are anti-parallel .

In this way we can define the ampere!!


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The rail gun:

Checkpoint #2

Is a device in which a magnetic force can accelerate (~10 6 g) a

projectile to a high speed (~10 km/s) in a short time (~ 1 ms).


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29-4 : Amperes law:

Ampere s law states that: the closed path line integral of B . ds around a circle concentric with the current equals m

oienc

.

As we learned in section 30.1, if a current carrying wire is

grasped with the right-hand with the thumb in the direction ofthe current, the fingers curl in the direction of B.

encoi sd B m


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Special cases:

The magnetic field outside a long (thick/ circular) straight wire:

The magnetic field inside a long (thick/ circular) straight wire withuniform current density:

B = mo i r/(2 p R 2) for r < R

B = mo i/(2 p r) for r > R


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Checkpoint #3

B = mo Js/2

What is the magnetic field created by an infinite uniform currentsheet J s, with a current i over a perpendicular length of the sheet Lsuch: i = J s/L?

Interaction:

Prove it!


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29-5: Solenoids and Toroids:

The field inside the solenoid is ~ uniform. The field between the turns

tend to cancel. The field outside the solenoid is weak!

The field of a solenoid is similar to that of a bar magnet!

An ideal solenoid is one for which the turns areclosely spaced and the length is long comparedto the radius.


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Applying Ampere s law to an ideal solenoid gives:

So, now we know how to create strong uniform magnetic fields!!

Why do we use superconducting coils?

B = mo (N/L) I = mo n I inside

B= 0 outside


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What is the magnetic field created by a toroid ?

Note that B is not everywhere constant inside the toroid!!

B = mo N i/(2 pr) inside

B = 0 outside


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29-6: A current carrying coil as a magnetic dipole:

B(z) = ( mo/2) i R 2/(R 2+z 2)3/2

We have already seen that if a circuit of magnetic dipole issituated in a magnetic field B, the circuit experiences a torque

produced such that:= x B

Moreover, one can show that for points on the central axis (take itto be the z-axis) of a single circular loop, a circulating current [i]

produces a magnetic field :


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Notice the similarity with the electric field and the electric dipole!!

Checkpoint #4

The circular loop current acts like a magnet:

B(z) = ( mo/2p) /z3

For points far from the loop (still on the z-axiz), this can becast in the form:

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Ch-29-Magnetic Fields Due to Currents