Question

Electric charge moves in straight line at rate of 3x10^-4 coulomb per minute, Calculate

the magnetic flux density at a distance of 20 cm from them.

Answer 1

Answer: The magnetic flux density at a distance of 20 cm from the moving charge is 2.5x10^-9 T

Step-by-step explanation:

To calculate the magnetic flux density, we need to use the formula for the magnetic field created by a moving charge:

`B=( u_(o) .q.v)/(4\pi r^(2) )`

where:

B is the magnetic flux density

`u_(o)` is the vacuum permeability constant

q is the charge of the moving charge (in coulombs)

v is the velocity of the moving charge (in m/s)

r is the distance from the moving charge (in meters)

In this problem, we have:

q = 3x10⁴ C (the rate at which the electric charge moves per minute)

v = ?

r = 0.2 m = 20 cm

To find the velocity (v), we need to convert the rate of electric charge to a current.

∴
`I = (q)/(t)`

where I is the electric current (in amperes) and t is the time (in seconds).

Since the rate is given in coulombs per minute, we need to convert the time to seconds:

t = 1 minute = 60 seconds

∴ I = q / t = (3x10^-4 C) / (60 s) = 5x10^-6 A

Now we can use the formula for the magnetic flux density:

`B=( u_(o) .q.v)/(4\pi r^(2) )`

Putting the values, we get:

`v = (B.4\pi r^2)/(u_(o).q )`

Simplifying, we get:

v = B * 2 * 10^3 m/s

Dividing both sides by 2*10^3 m/s, we get:

B = v / (2 * 10^3 m/s)

Plugging in the value of v, we get:

B = 5x10^-6 A / (2 * 10^3 m/s) = 2.5x10^-9 T

Therefore, the magnetic flux density at a distance of 20 cm from the moving charge is 2.5x10^-9 T.