Answer:
Explanation:
Given that,
The electron accelerated from rest through 2300V, then the Potential difference is 2300V
V = 2300V.
Magnetic field of 1.25T
B= 1.25T
Mass of electron
m = 9.11 ×10^-31kg
Charge of an electron
q = 1.602 ×10^-19C
Maximum magnetic force?
Maximum magnetic force can be determined using
Fmax = qvB
We don't know the velocity(v) but it can be determine using
Work done by potential = Kinectic energy
qV = ½mv²
Rearranging
v = √(2qV/m)
v = √(2×1.602×10^-19 × 2300/9.11×10^-31)
v = 2.84 × 10^7 m/s
Then,
Maximum Magnetic force
Fmax = qvB
Fmax=1.602×10^-19×2.84×10^7×1.25
Fmax = 5.7 × 10^-12 N
The magnitude of the maximum magnetic force is 5.7 × 10^-12 N
Answer:
[tex]F_{maximum} =5.68*10^{-12} \ \ N[/tex]
Explanation:
Electron acquire through kinetic energy is given as:
[tex]K = \frac{1}{2}mv^2\\ v^2 = \frac{2 \ K}{m}\\v = \sqrt{\frac{2 \ K}{m}}[/tex]
Given that :
Electron energy expanded = 2300 V
= 2300 eV
[tex]= 2300 *1.6*10^{-19} \ J[/tex]
= [tex]3.68*10^{-16} \ J[/tex]
Velocity acquired [tex]v = \sqrt{\frac{2 K}{m}}[/tex]
[tex]v = \sqrt{\frac{2 *3.66*10^{-16} }{9.11*10^{-31}}}[/tex]
[tex]v = 28423641.62[/tex]
[tex]v = 2.84 *10^7 \ m/s[/tex]
Force exerted by the magnetic field
[tex]F_B = Bvq sin \theta[/tex]
where θ = 90° and sin θ = sin 90 ° = 1
∴
[tex]F_{maximum} = 1.6*10^{-19} * 2.84*10^7 *1.25[/tex]
[tex]F_{maximum} =5.68*10^{-12} \ \ N[/tex]
