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:
the charged particle is electron..
Explanation:
electrons are the only charged particles which can be transferred by rubbing certain materials which generate electrostatic force ...
Hi,Find answers from Task 5
1.(X+4)+(X)+(X+4)+(X)=50cm
4x+8=50cm
4x=42
X=10.5cm
Length=10.5+4=14.5cm
Width=10.5cm
Area= length × width=(10.5/100) × (14.5/100) =0.0152m2
2. Volume of a sphere= 4/3 ×π×r³
4/3 ×π×r³=3.2×10^-6 m³
r³=3.2×10^-6 m³/1.33×π
r³=7.64134761e-7
r=0.00914m
Surface area of the blood drop= 4πr²
=4×3.142×0.00914×0.00914=0.00105m²
3.
Equation of an ideal gas = PV =n RT
Equation for pressure, = P= n RT/V
Equation for the volume of an ideal gas= V= n RT/P
If the volume of gas doubles ,V(new)= 2n RT/P
Equation for temperature of an ideal gas, T = PV/n R
If temperature of gas triples, T (new)= 3PV/n R
New Equation for Pressure, = n× R× (3PV/n R)/(2n RT/P)
Pressure factor increase= P(new)/P(old) ={ n× R× (3PV/n R)/(2n RT/P)}/{ n RT/V}
=3PV²/2n RT
