The Andromeda Galaxy (our nearest spiral neighbor) has spectral lines that show a blue shift. From this we may conclude that:
1 answer:
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Answer:
Current in the toroid will be
Explanation:
We have given number of winding in rectangular toroid N = 1500
Self inductance of toroid L = 0.06 H
Magnetic energy stored in toroid
We have to find the current in the toroid
Magnetic energy stored is equal to
So current in the toroid will be
Answer:
The value is
Explanation:
From the question we are told that
The value of charge on each three point charge is
The length of the sides of the equilateral triangle is
Generally the total potential energy is mathematically represented as
=>
Here k is coulomb constant with value
=>
The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
Given values:
Length of non-conducting rod, l = 1.20 m
Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C
Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C
Distance from point P of each rod, x = 60 cm = 0.60 m
Calculation of Net electric force exerted on point P:
Consider an electron released at point P, then the net electric force exerted will be given as:
F = e. E_net - ( 1 )
Step 1:
The net electric field value is given as:
E_net = E₁ cos Φ + E₂ cos Φ
= 2E₁ cos Φ -( 2 )
where, E₁ & E₂ are electric fields due to positive and negative rod
respectively.
Φ is phase angle
Step 2:
The electric field due to positive rod is given as:
E₁ = k (λ/r) - ( 3 )
where, k is Coulomb's force constant
λ is linear charge density
r is distance between point P and half of the rod.
Now, the linear charge density is given as:
λ = Charge/length = Q/x
The value of r is given as:
r = √x²-a²
where, x is length of rod
a is half length of rod
Applying values in above equation, we get:
r = √x²-(x/2)²
r = √(1.20 m)²-(1.20/2)²
= √1.08
= 1.04 m
Substituting all the determined values in equation 3 we get:
E₁ = k (λ/r)
= k [(Q/x)/r]
= k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 3:
Similarly, the electric field due to negative rod is given as:
E₂ = k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 4:
Consider equation 2:
E_net = 2E₁ cos Φ
From the figure we get the phase angle as:
Φ = tan⁻¹ (0.60 m/0.60 m)
= tan⁻¹ ( 1 )
= π/4
Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:
E_net = 2(1.803×10⁴ N/C) cos π/4
= 2(1.803×10⁴ N/C) (0.5)
= 18030 N/C
Step 5:
Consider equation 1 :
F = e. E_net
where, e is charge on an electron
Applying values in above equation we get:
F = (1.6 × 10⁻¹⁹ C)(18030 N/C)
= 2.885 × 10⁻¹⁵ N
Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
Learn more about electric force here:
<u>brainly.com/question/1634182</u>
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