The Magnetic Field Produced at the Center of the Circle

To determine the magnetic field produced at the center of the circular path of a helium nucleus, we can make use of the formula for the magnetic field of a moving charge in a circular path. The formula is given by:

B = μ₀I/2r

Where:
- B is the magnetic field
- μ₀ is the permeability of free space, which is a constant equal to 4π × 10^-7 T·m/A
- I is the current
- r is the radius of the circular path

Step 1: Determining the Current

In this case, the helium nucleus is moving in a circular path, which means it is experiencing centripetal force. This force is provided by the electromagnetic force, which is due to the motion of the charged particles. Therefore, we can equate the centripetal force to the electromagnetic force.

The centripetal force is given by:

Fc = mω²r

Where:
- Fc is the centripetal force
- m is the mass of the helium nucleus
- ω is the angular velocity
- r is the radius of the circular path

The electromagnetic force is given by:

Fem = qvB

Where:
- Fem is the electromagnetic force
- q is the charge of the helium nucleus
- v is the velocity of the helium nucleus
- B is the magnetic field

Since the centripetal force equals the electromagnetic force, we can equate the two equations:

mω²r = qvB

We can rearrange this equation to solve for the current:

I = qv/2πr

Step 2: Calculating the Magnetic Field

We are given that the radius of the circular path is 0.8 m and it takes 2 seconds to complete one revolution. From this information, we can determine the angular velocity:

ω = 2π/T

Where:
- ω is the angular velocity
- T is the time taken to complete one revolution

Substituting the given values:

ω = 2π/2 = π rad/s

The velocity can be calculated using the formula:

v = ωr

Substituting the values:

v = π × 0.8 = 2.51 m/s

Now, we can substitute the values of q, v, and r into the formula for current:

I = qv/2πr

I = (2e)(2.51)/(2π)(0.8) = 1.98 A

Finally, we can calculate the magnetic field using the formula:

B = μ₀I/2r

Substituting the values:

B = (4π × 10^-7)(1.98)/(2)(0.8) = 1.24 × 10^-6 T

Therefore, the magnetic field produced at the center of the circular path by the helium nucleus is 1.24 × 10^-6 Tesla.