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Chapter 4: Q30E (page 914)

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 0.0250 T. The ions next enter a second uniform magnetic field (B鈥�) oriented at right angles to their velocity. 40K contains 19 protons and 21 neutrons and has a mass of 6.64 * 10-26 kg. (a) What is the magnitude of the electric field in the velocity selector? (b) What must be the magnitude of B鈥� so that the ions will be bent into a semicircle of radius 12.5 cm?

Short Answer

Expert verified

a) The magnitude of the electric field is 112V/m

b) The magnitude of B鈥� 1.49脳10-2Tso that the ions will be bent into a semicircle of radius 12.5 cm

Step by step solution

01

Magnetic force and electric force

The magnetic force and the electric force are equal

FB鈫�=q(v鈫�脳B鈫�)AndFE鈫�=qE鈫�

Therefore,

FE鈫�=FB鈫�

02

Determine the magnitude of the electric field

(a)

The magnetic force and the electric force are equal

Therefore,

FB鈫�=q(v鈫�脳B鈫�)FE鈫�=qE鈫�FE鈫�=FB鈫�qvB=qEE=vBPutthevaluesintheequationE=vB=(4.50脳103m/s)(0.0250T)=112V/m

Therefore, the electric field is 112 V/m

03

Determine the magnitude of the magnetic field

(b)

The magnetic field is

B鈥�=mvqR=(6.64脳10鈭�26kg)(4.50脳103m/s)(1.60脳10鈭�19C)(0.125m)=1.49脳10鈭�2T

Therefore, the magnetic field is1.49脳10-2T

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