A singly ionized (one electron removed) 40 K atom passes through a velocity sele

Question

A singly ionized (one electron removed)40K atom passes through a velocity selector consisting of uniform perpendicular electric and magnetic fields. The selector is adjusted to allow ions having a speed of 4.50 km/s to pass through undeflected when the magnetic field is 2.50×102 T. The ions next enter a second uniform magnetic field ({ m B}') oriented at right angles to their velocity. 40K contains 19.0 protons and 21.0 neutrons and has a mass of 6.64×1026 kg.

Part A- What is the magnitude of the electric field in the velocity selector?

E=112 V/m

B'= ? T

Explanation / Answer

Given Charge on 40 K atom is q = 1.6*10-19 C Magnitude of magnetic field is   B = 2.5*10-2 T Velocity of ion is    v = 4.5 km / s                                   = ( 4.5 km /s ) ( 1000 m / 1 km )                                   = 4.5 *103 m/s Mass of potassium    m = 6.64*10-26 kg a) Magnitude of electric fiedl in velocity selector is                      E q = B q v                          E = B v                              = ( 2.5 *10-2 T ) ( 4.5*103 m/s )                              = 112.5 V / m ______________________________________________ b) Radius of semi circle is   r = 12.5 cm                                          = ( 12.5 cm ) ( 10-2 m / 1 cm )                                          = 12.5 *10-2 m Magnetic force on ion is equal to the centripetal force because it moves in semicircle                  m v 2 / r = B q v Magnitude of magnetic field is                           B = m v / q r                                = ( 6.64*10-26 kg ) ( 4.5*103 m/s ) / ( 1.6*10-19 C ) ( 12.5 *10-2 m )                                =0.01494 T                                 = 1.49*10-2 T

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