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 = Iπ
B cos 
For
= 0° →
U(Max) = MB cos(0) = MB = Iπ
B = 5 × π ×
× 3 ×
=
375 π x
.
For
= 90° →
U = MB cos (90) = 0
For
= 180° →
U(Min) = MB cos(0) = - MB = - Iπ
B = - 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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An unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.
<h3>
What does Newton's Second Law of Motion state?</h3>
It states that the force applied to the object is equal to the product of mass and acceleration.

- An object will accelerate when the net force applied on the object is more than zero or unbalanced.
- The acceleration is the change in the direction or speed of the object. To achieve acceleration the force must be greater in a direction.
- When force is greater in one the object move in that direction which is known as acceleration.
Therefore, an unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.
Learn more about Newton's Second Law of Motion.:
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Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
The attached file has a detailed solution of the given problem.
Answer:
E1 = 2996.667N/C E2 = 11237.5N/C
Explanation:
E1 = kQ1/r^2
=8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2
= 2996.667N/C
E2 = kQ2/r^2
= 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2
= 11237.5N/C
The direction are towards the point a
Answer:
14657.32 J
Explanation:
Given Parameters ;
Number of moles mono atomic gas A , n
1 = 4
.2 mol
Number of moles mono atomic gas B , n
2 = 3.2mol
Initial energy of gas A ,
K
A = 9500 J
Thermal energy given by gas A to gas B ,
Δ
K = 600
J
Gas constant
R =
8.314 J
/
molK
Let K
B be the initial energy of gas B.
Let T be the equilibrium temperature of the gas after mixing.
Then we can write the energy of gas A after mixing as
(3/2)n1RT = KA - ΔK
⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600
T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K
Energy of the gas B after mixing can be written as
(3/2)n2RT = KB + ΔK
⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600
⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600
⟹ KB = 14657.32 J