If you were unfortunate enough to be 5.5 mm away from such a lightning bolt, how large a magnetic field would you experience

The question is incomplete. Here is the complete question.

Lightning bolts can carry currents up to approximately 20kA. We can model such a current as the equivalent of a very long, straight wire.

(a) If you were unfortunate enough to be 5.5m away from such a lightning bolt, how large a magnetic field would you experience?

(b) How does this field compare to one you would experience by being 5.5cm from a long, straight household current of 5A?

(a) B = 7.27 x 10⁻⁴ T

              (b) Approximately 40 times higher than a household one.

Explanation: Using Biot-Savart Law, the magnetic field in a straight, long wire is given by

where:

(permeability of free space) = T.m/A

(a) If lightning bolt is compared to a long and straight wire, then magnetic field is

B = 7.27 x 10⁻⁴ T

The magnitude of magnetic field in a lightning bolt is 7.27 x 10⁻⁴ T

(b) Magnetic field in a household wire will be

B = 1.82 x 10⁻⁵ T

Comparing fields:

≈ 40

The filed for a lightning bolt is approximately 40 times higher than for a household wire.

the answer is neither ball has a charge.

because if they were both positively charged, they will repel each other.

if they were positively and negatively charged, they will seek.

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day