chapter28--PHY 131 JAB

chapter28--PHY 131 JAB - Chapter 28: Sources of magnetic...

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1 Chapter 28: Sources of magnetic field
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2 The magnetic force between two charges: q’ v’ q v v and v’ are parallel Experiments show that r F ˆ 2 v v r q q m r ˆ The proportionality constant is 2 2 7 0 /C Ns 10 1 4 - × = π μ μ 0 is called the permeability of free space §27.1 Magnetic field of a moving charge
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3 The unit vector points from the source of the magnetic field (q’) to the point of interest (here, the location of charge q). ( ) r v v F ˆ 4 2 0 m × × = r q q π μ When the velocities are not parallel, the previous result becomes The magnetic force of q’ on q. This result is valid only when v << c and v’<<c
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4 The force exerted on q by q’ is ( ) r v v F × × = ˆ 4 2 0 m r q q π μ Where the unit vector points from q’ to q. The shows that the force on q’ by q is not necessarily the same as the force on q by q’. Newton’s 3 rd law is violated!
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5 Example: Consider the situation shown. Let v = v’. q v q’ v’ d d r ˆ What is the force of q’ on q? ( ) r v v F ˆ 4 2 0 × × = r q q m π μ ( ) r v ˆ × Is into the page, so F m is to the top of the page. F m
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6 ° = 45 sin 4 2 0 m v v r q q F π μ The magnitude of the force of q’ on q is Example continued
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7 q v q’ v’ d d F m r ˆ 45 ° What is the force of q on q’? ( ) r v v F × × = ˆ 4 2 0 r q q m π μ ( ) r v × ˆ Is out of the page, so F m is to the left. Example continued
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8 ° = 45 sin 4 2 0 m v v r q q F π μ The magnitude of the force of q on q’ is The conclusion is the magnitudes of the forces are equal, but the forces are not in opposite directions. Newton’s 3 rd law fails. Example continued
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9 Recall that ( ) B v F × = q ( ) r v v F ˆ 4 2 0 × × = r q q π μ Compare to ( ) r v B ˆ 4 2 0 × = r q The B-field for a point charge
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10 Consider a current- carrying wire I P r ˆ θ d L r ( ) θ π μ sin 4 ˆ 4 2 0 2 0 v r q d dB r q d d = × = r v B §28.2 Magnetic field of a current element
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chapter28--PHY 131 JAB - Chapter 28: Sources of magnetic...

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