Thursday, March 25, 1999
Announcements:

Lecture notes:

• F = qv (cross product) B.... towards the center of the circle (centripetal force)
• F = ma_c = mv²/r.....  r = radius of the circle
• F = qvb
• Therefore,  qvB =  mv²/r
• v = qBr/ m
• v = wr ( from circular motion)
• v = 2pfr or   2pf = qB/m
• f = qB/ 2pm
• T = 1/f = 2pm / qB

• Let us assume  a charges particle has v at some arbitrary angle with respect to B as it enters the magnetic field, v can be resolved into v_x, v_y, and v_z.
• ( v_x is not affected by B, v_y and v_z   [magnitude = ( v_y² + v_z²)^1/2] are and show circular motion)
• The most general motion of a charged particle is a helix.
• See Figure 29-11c
• See Figure 29-12
• See also Problem 29-22

• F_E = qE (same direction)
• F_B = qv (cross product) B  ( force is perpendicular to B)
• F_E acts on a charge independent of whether v=0 or not zero
• When a charged particle moves under E, work is being done
• K.E. increases or decreases as v changes
• When a charged particle moves in B, DW = F_B ( dot product) ds = F_B (dot product) v dt = 0
• KE and speed do not change; direction of v does change

• F = qE + qv (cross product) B
• E is perpendicular to B

• B is into the paper
• E is Down
• qE is Up
• qv (cross product) B is Up
• If v= E/B, then qE + qv (cross product) B = F = 0
• Therefore, the charges move in a straight line as shown above.
• If v does not equal E/B, these charges will bend
• This can be used as a velocity or K.E. selector.

• In region I, with E and B, velocity, v, of ion is defined by E and B
• Ions move in a straight line
• In region II, no Electric Field, B¹ is the magnetic field.
• The ions will move in a circle of r
• You can show m/q = rBB¹/E

• No E, No B , the electrons will hit spot 1 on the screen
• Turn on E, no B, the electrons hit spot 2, vertical displacement = y from spot 1
• F = qE = ma_y
• a_y = qE/m
• y = (1/2)a_y t²        L = v_xt
• y= (1/2) (qE/m) (L/v_x)²
• y = (1/2) (qEL²/ mv_x²).... let this be equation 1
• Keep E on, turn on B and adjust B so that electrons hit spot 1 again.
• F_y = qE + q(v (cross product) B) = 0
• v_x = E/B.... let this be equation 2
• Combine Equations 1 and 2
• y = (1/2) (qEL²B²/ mE²) = (1/2) (qL²B²/ mE)
• or.... m/q = L²B²/ 2yE

• V_d = drift velocity of charges (same direction as i)
• n = number of charges per unit volume
• F_B = q (v_d (cross product) B)  multiplied by the number of charges = q (v_d (cross product) B)(nAL)
• Drift velocity is in the direction of the wire, therefore F_B = q (L (cross product) B)(nAL (absoulute value of v_d)
• Recall that i= qd/qt = nq(dx)A/dt = nqv_dA
• Therefore F_B = i (L (cross product) B), where L is the vector in the direction of i
• The above equation is true for a straight wire.
• What if the wire is not straight (direction of current not constant)?
• Divide the wire inot small segments
• dF_B = i (ds (cross product) B)
• If we want to know the total force on a long section of wire....
• F = i I(integral) ds (cross product) B
• The total force in any closed current loop in a magnetic field is zero.
• See Sample Problem 29-7 from page 714 of the text book.