Answer:
Explanation:
Given that,
Mass m = 6.64×10^-27kg
Charge q = 3.2×10^-19C
Potential difference V =2.45×10^6V
Magnetic field B =1.6T
The force in a magnetic field is given as Force = q•(V×B)
Since V and B are perpendicular i.e 90°
Force =q•V•BSin90
F=q•V•B
So we need to find the velocity
Then, K•E is equal to work done by charge I.e K•E=U
K•E =½mV²
K•E =½ ×6.64×10^-27 V²
K•E = 3.32×10^-27 V²
U = q•V
U = 3.2×10^-19 × 2.45×10^6
U =7.84×10^-13
Then, K•E = U
3.32×10^-27V² = 7.84×10^-13
V² = 7.84×10^-13 / 3.32×10^-27
V² = 2.36×10^14
V=√2.36×10^14
V = 1.537×10^7 m/s
So, applying this to force in magnetic field
F=q•V•B
F= 3.2×10^-19 × 1.537×10^7 ×1.6
F = 7.87×10^-12 N
Answer:
F = 7.86 * 10^-12 N
Explanation:
Given:-
- The mass of the particle , m = 6.64 × 10^-27 kg
- The charge of the particle, q = +3.20 × 10^-19 C
- The potential difference applied, ΔV = 2.45 × 10^6 V
- The strength of magnetic field, B = 1.60 T
Find:-
What is the magnitude of the magnetic force (F) exerted on the particle?
Solution:-
- To determine speed (v) of the accelerated particle under potential difference (ΔV) we will use the energy-work principle. Where work is done (U) by the applied potential difference to accelerate the particle (Δ K.E)
Δ K.E = U
0.5*m*v^2 = ΔV*q
v = √(2ΔV*q / m )
- The Lorentz force (F) exerted by the magnetic field (B) on the charge particle (q) with velocity (v) is given by:
F = q*v*B
F = q*√(2ΔV*q/m)*B
F = (+3.20 × 10^-19)*√(2(2.45 × 10^6)*(+3.20 × 10^-19) / 6.64 × 10^-27 )*(1.60)
F = 7.86 * 10^-12 N
