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 = 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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I think you're fishing for "temporary magnet" or something like that,
but I don't agree with it.
Credit card strips, refrigerator magnets, recording tape, bar magnets,
and big heavy horseshoe magnets are permanent magnets ... you don't
have to keep an electric current circulating around them to make them
magnetic.
But that doesn't mean that they stay magnetic no matter WHAT you do
to them. They can be DEmagnetized by being heated, dropped on the
floor, hit with a hammer, or in the presence of another, stronger magnet.
Answer:
The mass of the another block is 60 kg.
Explanation:
Given that,
Mass of block M= 100 kg
Height = 1.0 m
Time = 0.90 s
Let the mass of the other block is m.
We need to calculate the acceleration of each block
Using equation of motion
Put the value into the formula
We need to calculate the mass of the other block
Using newton's second law
The net force of the block M
....(I)
The net force of the block m
Put the value of T from equation (I)
Put the value into the formula
Hence, The mass of the another block is 60 kg.
It would be the first one and the third one
Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
Explanation:
From the question we are told that:
Electric field 
Distance 
At negative plate
Generally the equation for Velocity is mathematically given by
Therefore
