Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Answer:
Will be doubled.
Explanation:
For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:
where e₀ is a constant, the electric permittivity.
Now we can isolate V, the potential difference between the plates as:
Now, notice that the separation between the plates is in the numerator.
Thus, if we double the distance we will get a new potential difference V', such that:
So, if we double the distance between the plates, the potential difference will also be doubled.
Answer:
the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.
That is,
Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100
Explanation:
Let the intensity of light be represented by I
Let the distance of the star be d
I ∝ (1/d²)
I = k/d²
For the earth,
Iₑ = k/dₑ²
k = Iₑdₑ²
For the other planet, let intensity be Iₒ and distance be dₒ
Iₒ = k/dₒ²
But dₒ = 10dₑ
Iₒ = k/(10dₑ)²
Iₒ = k/100dₑ²
But k = Iₑdₑ²
Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100
Iₒ = Iₑ/100
Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.
