Answer:
(a) B = 2.85 × 
Tesla
(b) I = I = 0.285 A
Explanation:
a. The strength of magnetic field, B, in a solenoid is determined by;
r = 
⇒ B = 
Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field
B = 
= 
B = 2.85 × Tesla
b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;
B = μ I N/L
⇒ I = B ÷ μN/L
where B is the magnetic field, μ is the permeability of free space = 4.0 ×
Tm/A, N/L is the number of turns per length.
I = B ÷ μN/L
= 
I = 0.285 A
Answer:
10250 N/C leftwards
Explanation:
QA = 4 micro Coulomb
QB = - 5 micro Coulomb
AP = 6 m
BP = 2 m
A is origin, B is at 4 m and P is at 6 m .
The electric field due to charge QA at P is EA rightwards
The electric field due to charge QB at P is EB leftwards
The resultant electric field at P due the charges is given by
E = EB - EA
E = 11250 - 1000 = 10250 N/C leftwards
Explanation:
first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s
now using the 3rd law of motion
v^2=u^2+2as
0=9+2a2
a= -9/4m/s^2
now force=mass×accelration
=2kg×(-9/4)m/s^2
=4.5 N
4.5 newton force applied on the book!
✌️:)
Answer:
33.516 kJ
Explanation:
Potential energy is given by:
PE = mgh
Where m is the mass, g is acceleration due to gravity, and h is the height. In this case:
PE = 38kg x 9.8m/s^2 x 90m = 33516 kg m^2/s^2 = 33516 J = 33.516 kJ
Answer:
The distance from the charge is 3.35 m.
Explanation:
Given that,
Electric potential, V = 635 V
Magnitude of electric field, E = 189 N/C
We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :
d is the distance from charge
So, the distance from the charge is 3.35 m. Hence, this is the required solution.
