Answer:
7.1934 x 10^12 V/m.s
Explanation:
In order to do this exercise, you need to use the correct formula. Besides that, we need to identify our data.
First we have the radius of the plates which are circular, and it's 0.1 m. The current of the loop (I) is 2.0 A, and the radius of the loop is 0.2 m.
Now with this data, we use the next formula:
I = dE/dt Eo A
Where:
dE/dt = rate of electric field
Eo = constant of permittivity of free space
A = Area of circle
Solving for dE/dT:
dE/dt = I / Eo*A
Now, the area of the circle is A = πr²
A = 3.1416 * (0.1)² = 0.031416 m²
Now solving the electric field:
dE/dt = 2 / (8.85x10^-12 * 0.031416)
dE/dt = 7.1934 x 10^12 V/m.s
Use Newton’s second law: F=ma
m= 250kg. a= 750ms-2
F= 250 x 750 = 187 500N
<span>If the wire is then wound into a coil, the magnetic field is greatly intensified producing a static magnetic field around itself forming the shape of a bar magnet giving a distinct North and South pole.
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The net force of a pair of balanced forces is zero
Explanation:
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