Question

II. Practice calculations:

1. A current of 2.24A flows in a straight wire oriented along the x-axis so the current flows to the left. What is the
magnitude and direction of the magnetic field created by the wire at a point that is 0.022m above the wire? What is the
magnitude of the field at a point twice as far above the wire as point X? 10 times as far below the wire?

Answer 1

Hi there!

For a straight wire, we can use Ampère's Law to find its magnetic field at various distances from the wire.

Using the equation:

`\oint B\cdot dl = B \cdot l = \mu_0 i`

B = Magnetic field strength (T)
l = length of path of integration (m)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
i = Enclosed current (2.24 A)

This is a dot-product, so the cosine of the angle between the magnetic field and the path of integration is considered. However, since we are always tangential to the magnetic field, cos(0) = 1. We can simplify to B * l.

The length of the path of integration is equivalent to the circumference of a circle produced with a radius 'r' as a straight, long wire creates circular magnetic fields around the wire.

Therefore:

`C = 2\pi r\\\\B \cdot 2\pi r = \mu_0 i`

Solve for 'B'.

`B = (\mu_0i)/(2\pi r)`

1.

Plug in the given values and solve for the strength of the magnetic field at r = 0.022m.

`B = ((4\pi * 10^(-7))(2.24))/(2\pi (0.022)) = \boxed{20.364 \mu T}`

Using magnetic hand rules, a left-flowing (-x axis) current will result in a magnetic field of this strength INTO THE PAGE (or +y-axis if we assign the +/- z-axis to be up/down respectively) above the wire.

2.

Since B ∝ 1/R, if we double 'R', B will be halved.

Therefore:

`B_(2R) = (B_R)/(2) = (20.364 \mu T)/(2) = \boxed{10.182 \mu T}`

3.

The same logic applies. If we increase 'R' by 10x, B will decrease by 10x.

`B_(10R) = (B_R)/(10) = (20.364\mu T)/(10) = \boxed{2.0364 \mu T}`

However, since this point is BELOW the wire, the direction of the magnetic field differs. Using hand rules, the field would point OUT OF THE PAGE, or to the -y-axis.