Answer:
a)
b)
Explanation:
Given that
Charge = q
Magnetic filed = B
Radius = R
We know that kinetic energy KE
----------1
m v² = 2 .KE
The magnetic force F = q v B
Radial force
For uniform force these two forces should be equal
q v B R =m v²
q v B R = 2 .KE
Now put the velocity v in the equation
Answer:
The magnitude of the electric field is 0.1108 N/C
Explanation:
Given;
number of electrons, e = 8.05 x 10⁶
length of the wire, L = 1.03 m
distance of the field from the center of the wire, r = 0.201 m
Charge of the electron;
Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)
Q = 1.2896 x 10⁻¹² C
Linear charge density;
λ = Q / L
λ = (1.2896 x 10⁻¹² C) / (1.03 m)
λ = 1.252 x 10⁻¹² C/m
The magnitude of electric field at r = 0.201 m;
Therefore, the magnitude of the electric field is 0.1108 N/C
Answer:
hope it helps you.................
Answer:
2kg
Explanation:
100N = Force 50m/s²= Acceleration
Mass=?
the equation is;
Mass(kg)= Force(N) ÷ Acceleration (m/s²)
100÷50=2