Two 3 m long conductors are separated from each other by 5 mm and carry a current of 10 A dc. Calculate the force that one conductor exerts on the other.

0.012 N

Explanation:

The formula to apply is that adopted from the Ampere law which is;

F= μ* I₁*I₂*l /2πd   where

F is force that one conductor exerts on the other.

μ = magnetic permeability of free space = 4π×10⁻⁷ T. m/A

I₁ = current in conductor one=10 A

I₂ = current in conductor two= 10 A

l= length of conductor= 3 m

d= distance between the conductors = 5 mm = 0.005 m

Applying the values in the equation

F= 4π×10⁻⁷ *10*10*3 / 2π*0.005

F= 6 * 10⁻⁵ / 0.005

F=0.012 N

0.19 m/s and 0.45 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s