The magnitude of the magnetic field on the axis of the ring 5 cm from its center is 143 pT.
The radius of the nonconducting ring is R = 10 cm.
The ring is uniformly charged q = 10 μC.
The angular speed of the ring, ω = 20 rad/s
The ring is x = 5 cm from the center of the ring.
Now,
R = 10 cm = 0.1 m
q = 10.0 μC = 10 × 10⁻⁶ C
x = 5 cm = 0.05 m
The magnetic field on the axis of a current loop is given as:
B = [ μ₀ IR² ] / [4π(x² + R²)^{3/2} ]
Now, I = q / [2π/ω]
So, the magnitude of the magnetic field which is directed away from the center is:
B = [ μ₀ ωqR² ] / [4π(x² + R²)^{3/2} ]
B = [ μ₀ (200) (10 × 10⁻⁶) (0.1)² ] / [4π((0.05)² + (0.1)²)^{3/2} ]
B = 1.43 × 10⁻¹⁰ T
B = 143 pT
Learn more about the magnetic field here:
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Let's start from the Fluorine (F), which has 4 atoms on the right side: this means we must put a factor 4 in front of HF to balance the number of atoms of Fluorine.
Nowe we have 4 atoms of hydrogen, H, on the left side. Since we have a

on the right side (2 atoms), we must put a factor 2 in front of

, to balance the number of atoms of hydrogen.
So now the atoms of oxygen are also balanced (2 on both sides), as well as the atoms of Silicon (Si), so the balanced reaction is
Well first you can not go to the 2nd energy level without the first so the first takes up 2 electrons leaving you 5 electrons and the recommended amount is 8 electrons so you would have 5 electrons left
B. fungi
I hope this helps!
Answer:
The vapor pressure for a mist is 
Explanation:
From the question we are given that
The radius is 
The temperature is 
The vapor pressure of water 
The density of water is 
The surface tension of water is 
Generally the equation of that is mathematically represented as
Where P is the vapor pressure for mist
R is the ideal gas constant = 8.31
making P the subject in the formula


