What is the speed of a beam of electrons when under the simultaneous influence of E = 1.64×104 V/m B = 4.60×10−3 T Both fields are normal to the beam and to each other and produce no deflection of the electrons. When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s

Explanation:

Magnetic force(B) = 4.60×10^-3 T

Electric force(E) = 1.64×10^4 V/m

Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

v = (1.64×10^4) ÷ (4.60×10^-3)

v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

B = 4/3 ( i + j + k )

Explanation:

I = 3 k

L = 5 x 10⁻² m

F = - 0.2 i + 0.2 j

Let magnetic field be B

B = B₁ i + B₂ j + B₃ k

F = L ( I x B )

= 5 x 10⁻² 3 k x ( B₁ i + B₂ j + B₃ k )

=   15 x 10⁻² B₁ j  -   15 x 10⁻² B₂ i

Given

F = - 0.2 i + 0.2 j

equating equal terms

15 x 10⁻² B₂ = .2

15 x 10⁻² B₁ =  .2

B₁ = .02 / 15 x 10⁻² = 20 / 15 N

B₂ = .02 / 15 x 10⁻² = 20 / 15 N

Now wire is rotated so that current flows in positive x direction

I = 3 i

F = . 2 k

F = L ( I x B )

= 5 x 10⁻² x 3 i  x ( B₁ i + B₂ j + B₃ k )

= 15 x 10⁻² B₃ k

15 x 10⁻² B₃ = .2

B₃ = .2 /15 x 10⁻²

B₃ = 20 /15

B = B₁ i + B₂ j + B₃ k

20 /15 9 ( i + j + k )

B = 4/3 ( i + j + k )