What temperatures will you find the fourth state of matter?
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The range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x
90° 0
180° - 375 π x
We have current carrying wire in a form of a circle placed in a uniform magnetic field.
We have to the range of potential energies of the wire-field system for different orientations of the circle.
<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>The formula to calculate the magnetic potential energy is -
U = M.B = MB cos
where -
M is the Dipole Moment.
B is the Magnetic Field Intensity.
According to the question, we have -
U = M.B = MB cos
We can write M = IA (I is current and A is cross sectional Area)
U = IAB cos
U = IπB cos
For = 0° →
U(Max) = MB cos(0) = MB = IπB = 5 × π ×
× 3 ×
=
375 π x .
For = 90° →
U = MB cos (90) = 0
For = 180° →
U(Min) = MB cos(0) = - MB = - IπB = - 5 × π ×
× 3 ×
=
- 375 π x .
Hence, the range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x
90° 0
180° - 375 π x
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Answer:
KE = m
Explanation:
In the generation of energy from hydroelectric power station, the motion of water, and the turbines are paramount. The falling flowing water turns the blades of the turbine, which in-turn causes the movement of a coil within a strong magnetic field.
The motion of the coil which cuts the strong magnetic field induces current. Thus, the system generates electrical energy.
The equation that links kinetic energy (KE), mass (m) and speed (v) can be expressed as:
KE = m
Refer to the diagram shown below.
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω