A beam of electrons has radius r and contains 'n' electrons per cubic meter moving with velocity `vec(v)` along the beam (figure). Assume that a beam that is much longer than its diameter forms a cylinderically symmetric distribution of charge and current. The beam expands if the electric force exceeds the magnetic force. So long as the expansion is slow, the approximations of cylindrical symmetry and static fields remain valid.

What is net force on the electron at the edge of the beam?

Options

A. `(2n e^(2)r)/(varepsilon_(0)) (1-mu_(0)varepsilon_(0)v^(2))hat(r)`
B. `(n e^(2)r)/(2varepsilon_(0)) (1-mu_(0)varepsilon_(0)v^(2)) hat(r)`
C. `(4n e^(2)r)/(varepsilon_(0)) (1-mu_(0)varepsilon_(0)v^(2))hat(r)`
D. `(n e^(2)r)/(4varepsilon_(0)) (1-mu_(0)varepsilon_(0)v^(2))hat(r)`

Answer

B

Explanation

The total charge inside the cylinder is `Q=-n epir^(2)l`
Applying Gauss's law gives`phi_(E)=Q//in_(0)`
`2pirlE_(r)=(-n epir^(2)l)/in_(0) Rightarrow vec(E)=-(n er)/(2in_(0))hat(r)`
The circulation of `vec(B)` around the circle is `varepsilon = 2pirB_(0)`
The position direction for current is along the axis parallel to `vec(v)`
The current passing through the circle is
`I=-pir^(2)j=-pir^(2)nev`
Applying Ampere's law we obtain `varepsilon = mu_(0)I`
`2pirB_(0)=-mu_(0)pir^(2)nev Rightarrow vec(B) = 1/2mu_(0)nevr(-hat(theta))`
The total force acting on an electron is the sum of the electric and magnetic forces
`vec(F)=-e(vec(E)+vec(v)xxvec(B))=e((n er)/(2 in_(0))-mu_(0)(n erv^(2))/2)hat(r)=(n e^(2)r)/(2in_(0))(1-mu_(0)in_(0)v^(2))hat(r)]`