If a = 2.0 cm, b = 5.0 cm, and i = 20 a, 6.0 μt is the magnitude of the magnetic field at the point p, So the correct option is (a).
The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field.
= μ
i 
/ 
= μ i / 
As, is moving down and is moving up so,
- = (μ i / ) - [μ i / ]
- = μ i 24 / 
- = 
- = 5.98×
T ≈ 6μT
Therefore, 6.0 μt is the magnitude of the magnetic field .
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Answer:
ΔV = 20.1 V
Explanation:
As the positive plates are connected to each other, the capacitors are connected in parallel, so the total system load is the sum of the charges on each capacitor.
Q = Q₁ + Q₂
The charge on each capacitor is
Q₁ = C₁ ΔV₁
Q₁ = 24 10⁻⁶ 25
Q₁ = 6.00 10⁻⁴ C
Q₂ = C₂ ΔV₂
Q₂ = 13 10⁻⁶ 11
Q₂ = 1.43 10⁻⁴ C
The total set charge is
Q = (6 + 1.43) 10⁻⁴
Q = 7.43 10⁻⁴ C
The equivalent capacitance is
C_eq = C₁ + C₂
C_eq = (24 + 13) 10⁻⁶
C_eq = 37 10⁻⁶ F
Let's use the relationship to find the voltage
Q = C_eq ΔV
ΔV = Q / C_eq
ΔV = 7.43 10⁻⁴ / 37 10⁻⁶
ΔV = 2.008 10¹
ΔV = 20.1 V
This voltage is constant in the combination so it is also the voltage in capacitor C1
Infrared radiation extends from the red edge of the visible light
spectrum at 700 nanometers all the way to the shortest microwave
wavelength of 1 millimeter.
This range of wavelengths corresponds to a frequency range
of roughly 430 terrahertz at the edge of red light, down to the
highest microwave frequency, at 300 gigahertz.
Answer:
Springs are made of a variety of materials including copper and various forms of steel. The most common is high carbon steel as it is cheap, easy to work and a couple of other important properties.
Copper springs exist as well, but copper is more expensive than steel. However, in an environment where corrosion resistance is important, copper springs are a good alternative.
Explanation:
Explanation:
We have,
Slit separation, d = 0.329 mm
Distance between slit and screen, D = 2.20 m
The first bright fringe is formed at a distance off 2.9 mm from the center of the pattern. We need to find the wavelength.
For double slit experiment, the fringe width is given by :
is wavelength
So, wavelength is 433 nm.
