The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
Answer:
(a) Steel rod: 
Copper rod: 
(b) Steel rod: 
Copper rod: 
Explanation:
Length of each rod = 0.75 m
Diameter of each rod = 1.50 cm = 0.015 m
Tensile force exerted = 4000 N
(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as
Strain = 
where Y = Young modulus
F = Fore applied
A = Cross sectional area
For the steel rod:
Y = 200 000 000 000 
F = 4000N
A = 
(r = d/2 = 0.015/2 = 0.0075 m)
=> A = 
=> A = 0.000177 
∴ 
For the copper rod:
Y = 120 000 000 000 N/m²
F = 4000N
A = (r = d/2 = 0.015/2 = 0.0075 m)
=> A =
=> A = 0.000177
(b) We can find the elongation by multiplying the Strain by the original length of the rods:
Elongation = Strain * Length
For the steel rod:
Elongation = 
For the copper rod:
Elongation = 
Electric field strength = resistivity of copper x current density
where
p= 1.72 x 10^-8 <span>ohm meter
diameter = 2.05mm=.00205 m
current = 2.75 A
</span>get first the current density:
current density = current/ cross section area
find the cross section area
cross section area = pi.(d/2)^2;
cross section = 3.3 006x10-6 m^2
substitute the values
current density = 2.75A/3.3006x 10-6m^2
current density=35.55 x1 0^2 A/m^2
Electric field stregnth =1.72 x 10^-8 ohm meter x 35.55 x10^2 A/m^2
Electric field stregnth= 46.415 Volts/m
The electric field strength of copper is 46.415 V/m.
Answer: The answer to the question is 0.25 ohms
Explanation:
R = u x/A .......1
where u is chosen as the resistivity of
the rod, A is area of the rod and x is
chosen as the length of the rod.
Let R* be the resistance of the
lengths of the rod across the adjoints
Then R* = u1/4.......2
Comparing equation 1 and 2
R* = 1/4
=0.25ohms
