Q = magnitude of charge at each of the three locations A, B and C = 2 x 10⁻⁶ C
r₁ = distance of charge at origin from charge at B = 50 - 0 = 50 cm = 0.50 m
r₂ = distance of charge at origin from charge at C = 100 - 0 = 100 cm = 1 m
F₁ = magnitude of force by charge at B on charge at origin
F₂ = magnitude of force by charge at C on charge at origin
Magnitude of force by charge at B on charge at origin
inserting the values
F₁ = 0.144 N
Magnitude of force by charge at C on charge at origin
inserting the values
F₂ = 0.036 N
Net force on the charge at the origin is given as
F = F₁ + F₂
F = 0.144 + 0.036
F = 0.18 N
from the diagram , direction of net force is towards left or negative x-direction.
-- more mass involved
-- less distance between the two objects
Answer:
a) B = 10⁻¹ r
, b) B = 4 10⁻⁹ / r
, c) B=0
Explanation:
For this exercise let's use Ampere's law
∫ B. ds = μ₀ I
Where I is the current locked in the path. Let's take a closed path as a circle
ds = 2π dr
B 2π r = μ₀ I
B = μ₀ I / 2μ₀ r
Let's analyze several cases
a) r <Rw
Since the radius of the circumference is less than that of the wire, the current is less, let's use the concept of current density
j = I / A
For this case
j = I /π Rw² = I’/π r²
I’= I r² / Rw²
The magnetic field is
B = (μ₀/ 2π) r²/Rw² 1 / r
B = (μ₀ / 2π) r / Rw²
calculate
B = 4π 10⁻⁷ /2π r / 0.002²
B = 10⁻¹ r
b) in field between Rw <r <Rs
In this case the current enclosed in the total current
I = 0.02 A
B = μ₀/ 2π I / r
B = 4π 10⁻⁷ / 2π 0.02 / r
B = 4 10⁻⁹ / r
c) the field outside the coaxial Rs <r
In this case the waxed current is zero, so
B = 0
I = pressure amplitude given = 0.2 W/m²
dB = decibel reading
decibel reading from the pressure amplitude is given as
dB = 10 log₁₀ (I/10⁻¹²)
inserting the values in the above equation
dB = 10 log₁₀ (0.2/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹.10¹²)
dB = 10 log₁₀ (2 x 10¹²⁻¹)
dB = 10 log₁₀ (2 x 10¹¹)
dB = 113.01 db
hence the decibel reading comes out to be 113.01 db
Answer:
yes
Explanation:
because motion is relevant
