Chapter 28b - Magnetic fields (cont.) We found last time,...

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We found last time, that the force , , experienced by a charged particle , of charge q, moving with velocity through magnetic field , , is given by, Magnetic fields (cont.) B F G B G B Fq ( v B ) = × G G G This lets us control the path of charged particles using magnetic fields. v G Since electric fields also exert forces on charged particles, E E = G G We have two independent means of controlling the trajectory of charged particles. The net force on a charged particle in an electric and/or magnetic field can be expressed as, EB FF F q E q ( v B )q E ( v B ) ⎡⎤ =+= + ×= + × ⎣⎦ GG G G G G G G G
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The textbook describes (section 28-4) how J. J. Thomson exploited this control in his discovery of the electron . In an evacuated tube he used an cathode ray gun to fire what at the time were only understood to be charged particles through a region of crossed electric and magnetic fields.
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The crossed electric and magnetic fields were arranged to exert opposing forces on the particles so that the balance of these fields controlled the vertical position of the particles when they exited the field region, determining where they hit the phosphor coated screen .
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E L x y 2 1 2 qE L y mv ⎛⎞ = ⎜⎟ ⎝⎠ y In the September 2nd lecture we derived the vertical deflection of electrons moving through a vertical electric field between two charged parallel plates , arriving at the expression, v Recall that by the definition of the electric field as the force on a positive test charge the force on a negative charge is in a direction opposite the electric field so the electrons deflect upward. The upward force on the negative particles has magnitude given by, E Fq E = E field only
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+ A uniform magnetic field directed perpendicular to the electric field (into the page as indicated) generates a force on the negative particles that acts downwards . Its magnitude given by, L x y v B F qvBsin qvB = φ= (since φ = 90 o ) B field only
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+ E L x y 2 1 2 qE L y mv ⎛⎞ = ⎜⎟ ⎝⎠ y v Thompson’s brilliant procedure was to first turn on the electric field only and determine (from the position of the spot on the screen) the deflection y , related to the unknown incident speed , v , of the particles by our expression, Next he turned on the magnetic field and by adjusting its magnitude he brought the deflection to zero . At those particular fields the electric and magnetic forces were equal (but opposite) so he could equate the forces , qvB qE =
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22 11 qE L qE L y mv mE / B ⎛⎞ == ⎜⎟ ⎝⎠ E v B = This let him solve for the incident speed in terms of the measured electric and magnetic fields, Using this in the expression for the original deflection 2 qE L B y 2m E = mL B q2 y E = Since all the variables on the right were obtained from measurements Thompson had determined the ratio of the mass of the particles to their charge . He had thus demonstrated that “cathode rays”, consisted of unique sub–atomic particles that came to be known as electrons.
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This note was uploaded on 03/12/2010 for the course PHY PHY taught by Professor Mueller during the Spring '09 term at University of Florida.

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Chapter 28b - Magnetic fields (cont.) We found last time,...

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