The direction of the magnetic field in front of the wire closest to the student is on the left. The direction is found by the right-hand rule.
<h3>What is the right-hand rule?</h3>
The right-hand rule is a popular mnemonic for remembering how axes in three-dimensional space are oriented.
The fact that the three axes of three-dimensional space have two different orientations gives birth to the majority of the many left-hand and right-hand rules.
Using the right-hand rule, we can recall this diagram. Your thumb points in the direction of the magnetic force pushing on the moving charge
If you point your pointer finger in the direction of the positive charge and then your middle finger in the direction of the magnetic field.
To learn more about the right-hand rule refer to the link;
brainly.com/question/9750730
The electric force on the electron is opposite in direction to the electric field E. E points in the -y direction, so the electric force will point in the +y direction. The magnitude of the electric force is given by:
F = Eq
F = electric force, E = electric field strength, q = electron charge
We need to set up a magnetic field such that the magnetic force on the electron balances out the electric force. Since the electric force points in the +y direction, we need the magnetic force to point in the -y direction. Using the reversed right hand rule, the magnetic field must point in the -z direction for this to happen. Since the direction is perpendicular to the +x direction of the electron's velocity, the magnetic force is given by:
F = qvB
F = magnetic force, q = charge, v = velocity, B = magnetic field strength
The electric force must equal the magnetic force.
Eq = qvB
Do some algebra to isolate B:
E = vB
B = E/v
Let's solve for the electron's velocity. Its kinetic energy is given by:
KE = 0.5mv²
KE = kinetic energy, m = mass, v = velocity
Given values:
KE = 2.9keV = 4.6×10⁻¹⁶J
m = 9.1×10⁻³¹kg
Plug in and solve for v:
4.6×10⁻¹⁶ = 0.5(9.1×10⁻³¹)v²
v = 3.2×10⁷m/s
B = E/v
Given values:
E = 7500V/m
v = 3.2×10⁷m/s
Plug in and solve for B:
B = 7500/3.2×10⁷
B = 0.00023T
B = 0.23mT
Answer:
Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension
ex = бx/E
бx = Fx/A = Fx/π
Using both equation and solving for the modulus of elasticity E
E = бx/ex = Fx / πex
E = 
Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius
ey = 
(бy - v (бx + бz)) = 
бx
= 
= 
Finally
ey = Δr / r
Δr = ey * r = 10 * 
Δd = 2Δr = 
Explanation:
