A long straight conducting rod carries a current I with a non-uniform current density J = ar2, and has a radius R. The value of the constant is 28.5 A/mm4 and the radius of the rod is 5.20 mm. Determine the magnitude of the magnetic field at the following points.(a) r1 = R/2(b) r2 = 2R

Following are the solution to this question:

Explanation:

Given value:

Current density

value of the constant is=  

radius = 5.20 mm

magnetic permeability

calculating the area element for the straight circular conduction rod :

Calculating the current, which is carrying in rod

Calculating the above equation with the limit value that is 0 to r.

The calculated current value which is carried by the rod is

In option (a)

calculating the magnitude of the magnetic field at the point

Substituting the above value:

Thus, the magnitude of the magnetic field at the point is

In option (b)

In this, we calculate the magnitude of the magnetic field at the point

Substituting the values  

Thus, the magnitude of the magnetic field at the point is

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