Question

A point charge is moving with speed 2 × 107 m/s parallel to the x axis along strait line at y=2 m. At t = 0, the charge is at x = 0 m. The magnitude of the magnetic field at x = 4 m is B0 at the origin. The magnitude of the magnetic field at at origin when t = 0.15 μs is?

Answer 1

The magnitude of the magnetic field at the origin when t = 0.15 μs can be found using the formula for the magnetic field produced by a moving charge.

Step-by-step explanation:

The magnitude of the magnetic field at the origin when t = 0.15 μs can be found using the formula for the magnetic field produced by a moving charge:

B = (μ₀Iv)/(2πr)

Where B is the magnetic field, μ₀ is the permeability of free space, I is the current, v is the velocity of the charge, and r is the distance from the charge to the point where we want to find the magnetic field. In this case, the charge is moving parallel to the x-axis, so the current can be calculated using the formula:

I = (Qv)/L

Where Q is the charge and L is the length of the moving charge. Plugging in the values and solving the equations will give the magnitude of the magnetic field at x = 4 m when t = 0.15 μs.

Answer 2

`B_2  = 4B_1`

Step-by-step explanation:

initial distance of point from the electron is r_1 = 4 m

distance moved by electron is x

x = vt

`x = 2*10^(7) *0.1*10^(-6)`

x = 2 m

new distance of point from electron is r_2 = 2m

field due to moving charge is

`B = ( \mu_o 4vr)/(4 \pi r^3)`

CHARGE IS INVERSELY PROPOTIONAL TO DISTANCE FROM ELECTRON.

therefore
`(B_1)/(B_2) = [(r_2)/(r_1)]^2`

`= [(2)/(4)]^2`

`B_2  = 4B_1`