A small compass is held horizontally, the center of its needle a distance of 0.460 m directly north of a long wire that is perpendicular to the earth's surface. When there is no current in the wire, the compass needle points due north, which is the direction of the horizontal component of the earth's magnetic field at that location. This component is parallel to the earth's surface. When the current in the wire is 19.0 A, the needle points 11.0° east of north. (a) What direction does the current in the wire flow, toward or away from the earth's surface? (b) What is the magnitude of the horizontal component of the earth's magnetic field at the location of the compass?
The field of the wire has a magnitude B(wire) = µ₀•I/(2πd) = …
B(wire)/B(hor) = tan11⁰ ==>
B(hor) = B(wire)/ tan11⁰ =
To answer your question, we can use the right-hand rule and apply the concept of the magnetic field produced by a current-carrying wire.
(a) First, let's determine the direction of the magnetic field produced by the current in the wire. Use your right-hand, extend your thumb in the direction of the current flow (from positive to negative). Your fingers will curl in the direction of the magnetic field lines.
In this case, since the compass needle is deflected to the east of north, the direction of the magnetic field produced by the wire must be toward the earth's surface. So the current in the wire flows away from the earth's surface.
(b) To find the magnitude of the horizontal component of the earth's magnetic field, we can use the tangent function and the angle by which the compass needle is deflected.
Let B_e be the magnitude of the horizontal component of the earth's magnetic field, and B_w be the magnitude of the magnetic field produced by the current in the wire.
The tangent of the angle by which the compass needle is deflected is given by:
tan(11°) = B_w / B_e
Rearranging the equation to solve for B_e, we have:
B_e = B_w / tan(11°)
Now, let's calculate B_e. We know the current in the wire is 19.0 A.
Using the right-hand rule, we can determine the direction of the magnetic field produced by the wire is perpendicular to the plane formed by the wire and the compass needle. Thus, it is perpendicular to the horizontal direction of the earth's magnetic field.
Given that the compass needle points due north when there is no current, the magnetic field produced by the wire cancels out the horizontal component of the earth's magnetic field.
Therefore, the magnitude of the horizontal component of the earth's magnetic field at the location of the compass is:
B_e = 0 / tan(11°) = 0
So, the magnitude of the horizontal component of the earth's magnetic field is zero at the location of the compass.
To determine the direction in which the current in the wire flows, we can use the right-hand rule for electromagnetism.
The right-hand rule states that if your right-hand thumb points in the direction of the current flow in a wire, then your curled fingers will point in the direction of the magnetic field surrounding the wire. In this case, since the needle of the compass points east of north, which means there is a magnetic field present due to the current in the wire.
To apply the right-hand rule, hold your right hand with your thumb pointing east (the direction the needle points), then your fingers will curl in the direction opposite to the current flow. In this case, the current in the wire flows towards the earth's surface.
Now, let's calculate the magnitude of the horizontal component of the earth's magnetic field using the given information.
We know that when there is no current in the wire, the compass needle points due north, which means the magnetic field due to the earth's horizontal component cancels out the magnetic field due to the current in the wire.
Given:
Distance between the needle's center and the wire, d = 0.460 m
Current in the wire, I = 19.0 A
Angle the needle points east of north, θ = 11.0°
The magnetic field due to the wire can be calculated using Ampere's Law:
B = (μ₀ * I) / (2π * d)
Where:
B is the magnetic field
μ₀ is the permeability of free space (4π x 10^-7 Tm/A)
I is the current in the wire
d is the distance between the wire and the compass needle's center
Substituting the given values:
B = (4π × 10^-7 Tm/A * 19.0 A) / (2π * 0.460 m)
Simplifying, we get:
B = (4π × 10^-7 x 19.0) / (2π x 0.460) T
B = 0.164 T
Therefore, the magnitude of the horizontal component of the earth's magnetic field at the location of the compass is 0.164 Tesla.