Answer:
dx/Dt x B . x =0
Explanation:
Let's calculate the work and the magnetic force, the expression for magnetic force is
F = qv x B
Bold indicate vector quantities, the expression for the job is
W = F. X
Let's replace in this equation
W = q v x B . X
The definition of speed is
v = dX / dt
With what work is left
W = q dX / dt x B . X
As we can see the vector product gives us a vector perpendicular to dX and its scalar product by X of zero
Second part
The speed a vector and although the magnitude is constant the change of direction implies a change in the speed.
Let's calculate the magnitudes of speed (speed)
F = qv B sin θ
F = ma
q v B sin θ = ma
a = qvB / m senT
This acceleration is perpendicular to the magnetic field and the velocity, so it does not change if magnitude but its direction, it is directed to the center of the circle.
| v | = q vB/m sin θ
Answer:
10 newtons, because the gravitaional force willbe stronger the closer it gets.
Answer:
220 A
Explanation:
The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.
So, F = BI₁L
F = (μ₀I₂/2πd)I₁L
F = μ₀I₁I₂L/2πd
Given that the current in the rods are the same, I₁ = I₂ = I
So,
F = μ₀I²L/2πd
Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²
So, F = W
μ₀I²L/2πd = mg
making I subject of the formula, we have
I² = 2πdmg/μ₀L
I = √(2πdmg/μ₀L)
substituting the values of the variables into the equation, we have
I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])
I = √(0.0049 × 10⁷kgm²/s²H)
I = √(0.049 × 10⁶kgm²/s²H)
I = 0.22 × 10³ A
I = 220 A
