An ion source is producing 6Li ions, which have charge +e and mass 9.99 10-27 kg

Question

An ion source is producing 6Li ions, which have charge +e and mass 9.99 10-27 kg. The ions are accelerated by a potential difference of 10 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude B = 1.2 T.

(a) If the 6Li ions are to pass through undeflected in the region with both E and B, what is the net force on the particle?

(b) Assume that B points into the page. What direction does E have to point for the ions to pass through undeflected? (c) What is the velocity of the particle after being accelerated by the potential difference?

(d) Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the 6Li ions to pass through undeflected.

Explanation / Answer

Given that

An ion source is producing 6Li ions, which have charge +e and mass (m) = 9.99 10-27 kg.

The charge of the electron (q) =1.6*10-19C

The ions are accelerated by a potential difference of (V) = 10 kV=10000V

which there is a uniform vertical magnetic field of magnitude (B) = 1.2 T

The mass of the electron is given by (m) =9.11*10-31kg

The force acting on the particle is given by F =qvB or F =qE

The electron acclerating in a potenial diffrenece is given by

W =qV =(1/2/)mv2

v =Sqrt(2qV/m) =Sqrt(2*1.6*10-19C*10000/(9.11*10-31kg)=59.267*106m/s =5.9267*107m/s

We know that the strength of theelectric field is given by

E =vB =(5.9267*107m/s)(1.2)=7.112*107V/m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.