Answer
current density is given by
where J = σ E
now,
a) Electric field strength at the inner surface of an iron cylinder
a = 0.5 cm = 0.005 m b = 2.3 = 0.023
L = 10 cm = 0.1 m I = 27 A
E = 8.59 x 10⁻⁴ V/m
b) Electric field strength at the outer surface of an iron cylinder
a = 0.5 cm = 0.005 m b = 2.3 = 0.023
L = 10 cm = 0.1 m I = 27 A
E =1.87 x 10⁻⁴ V/m
The strength of the electromagnet depends on how many coils you wrap round and how high the voltage is. ... N Increasing the number of coils, which adds more field lines and makes the electromagnet stronger. This is the magnetic field around a piece of wire, compared to a magnetic field on a loop or solenoid it is weak.
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Answer:
110 m/s²
Explanation:
Given:
v₀ = 0 m/s
Δx = 1.08 m
t = 0.14 s
Find: a
Δx = v₀ t + ½ at²
1.08 m = (0 m/s) (0.14 s) + ½ a (0.14 s)²
a = 110 m/s²
Answer:
A POSTAL PACKAGE IS PUSHED BY A HORIZONTAL TABLE WITH A FORCE F = 40N. PACKAGE MOVES ON A DISTANCE OF 1.5M WHICH MECHANICAL WORK PERFORMS THIS FORCE
PT TEST
Answer:
he correct answer is V = ER
Explanation:
In this exercise they give us the electric field on the surface of the sphere and ask us about the electric potential, the two quantities are related
ΔV = ∫ E.ds
where E is the elective field and normal displacement vector.
Since E is radial in a spray the displacement vector is also radial, the dot product e reduces to the algebraic product.
ΔV = ∫ E ds
ΔV = E s
since s is in the direction of the radii its value on the surface of the spheres s = R
ΔV = E R
checking the correct answer is V = ER
