At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field t
1 answer:
You might be interested in
Answer:
the electric field at point A is
E = 5.5 ×10¹³N/C(-x direction)
Explanation:
given
electrostatics constant k = 9.0×10⁹
charge q = 31.7mC= 31.7×10⁻³C
distance r = 2.80mm
distance from midpoint to point A = 2.00mm
attached is the diagram of the solution, describing the position of the charge
note x = r/2, where x is the distance from midpoint of r to the particle
using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m
the electric field at point A is given as
vector <em>E </em>= 2E×cos θ( -x direction)
recall E =kq/x²
where k is the electrostatics constant = 1/4πε₀
where ε₀ is permittivity of free space
therefore using E =2{kq/x²}cosθ
∴cosθ = adjacent/hypotenuse
cosθ=1.40/2.44
E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)
E=2{4.79×10¹³}×(0.574)(-x)
E = 2×2.75 ×10¹³N/C(-x direction)
Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)
Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 × m/s
uniform electric field E = 1.18 × N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 × kg ×a = 1.18 ×
× 1.602 ×
a = 20.75 × m/s²
so acceleration is 20.75 × m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 × = 0 + 20.75 ×
(t)
t = 11.80 × s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 × s
time will elapse before it return to its staring point is 23.6 ns