Answer:
The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Explanation:
Given;
radius of the circular loop of wire = 0.5 m
current in circular loop of wire = 2 A
strength of magnetic field in the wire = 0.3 T
τ = μ x Bsinθ
where;
τ is the magnitude of the magnetic torque
μ is the dipole moment of the magnetic field
θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90
μ = IA
where;
I is current in circular loop of wire
A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²
μ = 2 x 0.7885 = 1.571 A.m²
τ = μ x Bsinθ = 1.571 x 0.3 sin(90)
τ = 0.4713 J
Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Answer:
Explanation:
350 N force stretches the spring by 30 cm
spring constant K = 350 / 0.30 = (350 / 0.3) N / m
To calculate work done by a spring force we proceed as follows
spring force when the spring is stretched by x = Kx
This force is variable so work done by it can be calculated by integration
Work done by it in stretching from x₁ to x₂
W = ∫ F dx
= ∫ Kx dx with limit from x₁ to x ₂
= 1/2 K ( x₂² - x₁² )
Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m
Work done
= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )
= 227.50 J
Answer:
374.39 J/K
Explanation:
Entropy: This can be defined as the degree of disorder or randomness of a substance.
The S.I unit of entropy is J/K
ΔS = ΔH/T ..................................... Equation 1
Where ΔS = entropy change, ΔH = Heat change, T = temperature.
ΔH = cm................................... Equation 2
Where,
c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.
Substitute into equation 2
ΔH = 333000×0.3071
ΔH = 102264.3 J.
Also, T = 273.15 K
Substitute into equation 1
ΔS = 102264.3/273.15
ΔS = 374.39 J/K
Thus, The change in entropy = 374.39 J/K
