The main requirement for a good conductor of electricity is to have a lot of valence electrons. Valence electrons are the electrons of the outer shells of atoms not bound with other atoms (for example through covalent bounds). These electrons are "free to escape" as soon as an electric field with enough intensity is applied to the material, and therefore these electrons will be free to move in the material producing an electric current.
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
Answer:
Electromagnets are special types of magnets that are made by passing current through coils of wire. To make an electromagnet, the minimum requirements are:
1. A nail (usually made of iron, steel or zinc)
2. Dry cell batteries
3. Wire (Usually copper wire)
Other things could be:
1. Electric tape to hold both ends of the wire properly at the battery terminals.
2. Scissors to cut the wire into desired length.
3. Iron fillings for testing purposes.
Answer:
The angular speed of the object is 0.0281 rad/s
The linear speed of the object is 0.169 ft/s
Explanation:
Given;
radius of the circle, r = 6 ft
time of motion of the object around the circle, t = 80 s
central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad
The angular speed of the object is calculated as;
The linear speed of the object is calculated as;
v = ωr
v = 0.0281 rad/s x 6ft
v = 0.169 ft/s
Answer:
Distance of the point where electric filed is 2.45 N/C is 1.06 m
Explanation:
We have given charge per unit length, that is liner charge density 
Electric field E = 2.45 N/C
We have to find the distance at which electric field is 2.45 N/C
We know that electric field due to linear charge is equal to
, here 
is linear charge density and r is distance of the point where we have to find the electric field
So 
r = 1.06 m
So distance of the point where electric filed is 2.45 N/C is 1.06 m
