Number 12 gauge wire, commonly used in household wiring, is 2.053mm in diameter and can safely carry up to 20A. For a wire carrying the maximum current, find the magnetic field strength:

The magnetic field strength is 3.9 x 10⁻³ T

Explanation:

The equation for the magnetic field strength produced by a long straight current-carrying wire is written as;

Where;

μ₀ is constant = 4π x 10⁻⁷ N/A

I is the maximum current = 20 A

r is the radius of the wire = 2.053/2 = 1.0265 mm = 1.0265 x 10⁻³ m

Substitute in this values into the equation above;

Therefore, the magnetic field strength is 3.9 x 10⁻³ T

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s