Answer:
a
When 
b
When 
![B = [\frac{\mu_o * I }{ 2 \pi R^2} ]* r](https://tex.z-dn.net/?f=B%20%3D%20%20%5B%5Cfrac%7B%5Cmu_o%20%2A%20%20I%20%7D%7B%202%20%5Cpi%20R%5E2%7D%20%5D%2A%20r)
Explanation:
From the question we are told that
The radius is R
The current is I
The distance from the center
Ampere's law is mathematically represented as
![B[2 \pi r] = \mu_o * \frac{I r^2 }{R^2 }](https://tex.z-dn.net/?f=B%5B2%20%5Cpi%20r%5D%20%20%3D%20%20%5Cmu_o%20%20%2A%20%20%5Cfrac%7BI%20r%5E2%20%20%7D%7BR%5E2%20%7D)
When
=>
But when
Answer:
2.5m/s²
Explanation:
Given parameters:
Mass of car = 2000kg
Force produced by the car = 5000N
Unknown:
Acceleration of the car = ?
Solution:
According to Newton's second law of motion, Force is a product of mass and acceleration.
Force = mass x acceleration
Now, insert the parameters and find the unknown;
5000 = 2000 x acceleration
Acceleration = 
= 2.5m/s²
Answer:
a = v²/r
Explanation:
The acceleration of a body moving in a circular path is known as the centripetal acceleration. This is the acceleration of a body that keeps the body within the circular path. It is written in terms of the linear velocity v and the radius of the circle of rotation as shown;
a = v²/r where
v is the linear velocity
r is the radius
a is the centripetal acceleration
Answer:
<em>The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.</em>
Explanation:
Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.
Answer:
Explanation:
wave length of light λ = 623 x 10⁻⁹ m .
Distance of screen D = 76.5 x 10⁻² m
width of slit = d
Distance on the screen between the second order minimum and the central maximum = 2 λ D / d
1.11 x 10⁻² = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d
d = ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²
= 85872.97 x 10⁻⁹
= 85.87297 x 10⁻⁶
= 85.87 μm
width a of the slit is = 85.87 μm
