A proton travels at a speed of 7.20×107 m/s along the x-axis. What is the magnetic field at the point (0.00 cm, -7.00 cm, 0.00 cm) when the proton passes through the origin?

To solve this problem, we can use the formula for the magnetic field created by a moving charge, which is given by the Biot-Savart law:

B = (μ0/4π) * (q*v/r²) * r̂

where:

• B is the magnetic field,
• μ0 is the permeability of free space (4π x 10^-7 T m/A),
• q is the charge of the proton (1.6 x 10^-19 C),
• v is the speed of the proton (7.20 x 10^7 m/s),
• r is the distance from the proton to the point where we want to find the magnetic field, and
• r̂ is the unit vector pointing from the proton to the point where we want to find the magnetic field.

First, we need to find r. The proton is at the origin (0,0,0) and the point where we want to find the magnetic field is (0,-7 cm,0). So, the distance r is 7 cm = 0.07 m.

Next, we need to find r̂. Since the proton is moving along the x-axis and the point where we want to find the magnetic field is below the x-axis, r̂ points in the negative y-direction. So, r̂ = -ĵ.

Now we can plug in the values into the formula:

B = (4π x 10^-7 T m/A / 4π) * (1.6 x 10^-19 C * 7.20 x 10^7 m/s / 0.07 m²) * -ĵ

After simplifying, we find:

B = -1.3 x 10^-19 T ĵ

So, the magnetic field at the point (0,-7 cm,0) when the proton passes through the origin is -1.3 x 10^-19 T in the negative y-direction.