Question

1) Jumper cables used to start a stalled vehicle oen carry a 70-A current. How strong is the magnetic field 5.0 cm from one cable? Compare to the Earth’s magnetic field (5.0 × 10−5 T).

2) A power line carries a current of 100 A west along the tops of 10-m-high poles. (a) What is the magnitude and direction of the magnetic field produced by this wire at the ground directly below? How does this compare with the Earth’s magnetic eld of about 1 2 G? (b) Where would the wire’s magnetic field cancel the Earth’s field?

Answer 1

Solution (1)

Let us go to the basics first.

Just apply the magnetic field formula:

B = (Uo)I / 2pi*r

B = (4*pi*10^-7 * 70A) / (2pi * 0.05m)

B = 2.8 x 10^-4 Tesla (Answer1)

Compare to the Earth’s magnetic field (5.0 × 10−5 T).:

The above calculated magnetic field is :

(2.8x10^-4T / 5x10^-5T) *100 = 560% or 5.6 times greater than Earth's magnetic field (Answer1)

Solution (2):

First, the magnitude of the field is given by

B = 2 × 10⁻⁷ (100 / 10) T = 2 µT. (Answer 2a)

The direction is given by the right hand rule:

Holding your thumb in the west direction, the bottom of the fingers will point to the South.

Thus the magnetic field is 2 µT pointing South. (Answer 2a)

The Earth’s field is about 50 µT (1/2 G given) pointing to the North. Thus this power line produces a magnetic field that is small compared to the Earth’s field:

B / B_Earth = 2 / 50 = 4 %. The power line diminishes the natural magnetic field at the surface by 4 %.

(Answer 2a)

B = 2 × 10⁻⁷ (100 / h) T

=>50 µT = 2 × 10⁻⁷ (100 / h) T

=>50 µ = 2 × 10⁻⁷ (100 / h)

=>50*10^-6 = 2 × 10⁻⁷ (100 / h)

=>h = 0.4 m (Answer 2b)

Thanks!!!