Answer:
The inductance of solenoid A is twice that of solenoid B
Explanation:
The inductance of a solenoid L is given by
L = μ₀n²Al where n = turns density, A = cross-sectional area of solenoid and l = length of solenoid.
Given that d₁ = 2d₂ and l₂ = 2l₁ and d₁ and d₂ are diameters of solenoids A and B respectively. Also, l₁ and l₂ are lengths of solenoids A and B respectively.
Since we have a cylindrical solenoid, the cross-section is a circle. So, A = πd²/4.
Let L₁ and L₂ be the inductances of solenoids A and B respectively.
So L₁ = μ₀n²A₁l₁ = μ₀n²πd₁²l₁/4
L₂ = μ₀n²A₂l₂ = μ₀n²πd₂²l₂/4
Since d₁ = 2d₂ and l₂ = 2l₁, sub
L₁/L₂ = μ₀n²πd₁²l₁/4 ÷ μ₀n²πd₂²l₂/4 = d₁²/d₂² × l₁/l₂ = (2d₂)²/d₂² × l₁/2l₁ = 4d₂²/d₂² × l₁/2l₁ = 4 × 1/2 = 2
L₁/L₂ = 2
L₁ = 2L₂
So, the inductance of solenoid A is twice that of solenoid B
Explanation:
It is given that,
Mass of platinum bar, m = 750 g
Length of the bar, l = 5 cm
Breadth of the bar, b = 4 cm
Width of the bar, h = 1.5 cm
Volume of the bar, 
We need to find the density of the platinum bar. The density of any substance is given by :
So, the density of the platinum bar 
. Hence, this is the required solution.
A wind turbine turns wind energy into electricity using the aerodynamic force
Measurement=<span>the size, length, or amount of something
</span><span>unit=a quantity chosen as a standard in terms of which other quantities may be expressed</span>
The portion of the flux leaves the curved surface of the cylinder is 60%.
<h3 /><h3>What are electrons?</h3>
The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.
If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.
If 20% of the flux leave from one end, then another 20% will leave from another end.
So, the net flux through curved surface is
100 -20 -20 = 60%
Thus, the total flux leaves the curved surface of the cylinder is 60%
Learn more about electrons.
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