The magnetic field at the center of the arc is 4 × 10^(-4) T.
To find the answer, we need to know about the magnetic field due to a circular arc.
<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
- According to Biot savert's law, magnetic field at the center of a circular arc is
- B=(μ₀ I/4π)× (arc/radius²)
- As arc is given as angle × radius, so
B=( μ₀I/4π)×(angle/radius)
<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>
B=(μ₀ I/4π)× (0.9/0.006)
= (10^(-7)× 26.9)× (0.9/0.006)
= 4 × 10^(-4) T
Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.
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Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² = 
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and 
for 
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
The tensile stress of the wire supporting 2 kg mass is determined as 6.1 x 10⁷ N/m².
<h3>
Tensile stress of the wire</h3>
The tensile stress of the wire is calculated as follows;
σ = F/A
where;
A = πr² = πD²/4
where;
A = π x (0.64 x 10⁻³)²/4
A = 3.22 x 10⁻⁷ m²
σ = F/A = (mg)/A = (2 x 9.8)/( 3.22 x 10⁻⁷)
σ = 6.1 x 10⁷ N/m²
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Answer:
option d and b..............
Answer:
what do you need help with?
Explanation:
I would look up the periodic table of elements
hope this helps!
