Magnetic force around a current-carrying wire

Direction of the magnetic force around a current-carrying wireA straight wire carrying an electric current creates a magnetic field around the wire in concentric circles. This magnetic field can exert a magnetic force on nearby moving charged particles. Using the right-hand rule in the illustration at right, you can see that a positively charged particle moving in the same direction as the electric current will feel a magnetic force toward the wire. Likewise, a positively charged particle moving in the opposite direction as the electric current will feel a magnetic force away
from the wire
. Read the text aloud Show Vectors in and out of the page
Strength of the magnetic force around a current-carrying wireIf the direction of the magnetic force is given by the right-hand rule, what is the magnitude or strength of the magnetic field and force? The magnitude of the magnetic field around a straight, current-carrying wire is given by equation (19.6). The magnetic field B is stronger for a larger electric current I and becomes weaker as the distance r from the wire increases. Read the text aloud
(19.6) B= μ 0 I 2πr
B  = magnetic field (T)
μ0  = permeability constant = 4π×10−7 T m/A
I  = current (A)
r  = distance from wire (m)
Magnetic field
around a
current-carrying wire
Once the magnitude of the magnetic field is known, the magnetic force can be calculated by using equation (19.4), or FB = qvB sin θ, where θ is the angle between the magnetic field and velocity vectors. Read the text aloud
In equation (19.6), most of the quantities are familiar: magnetic field B, measured in teslas (T); electric current I, in amperes; and distance r from the wire , in meters. What is μ0? It is called the permeability of free space, which is a physical constant describing how effectively a magnetic field can be established in a vacuum. (It is sometimes called the vacuum permeability.) But all you really need to know to use equation (19.6) is that μ0 is a physical constant equal to 4π×10−7 T m/A! Read the text aloud Show Vacuum permeability
Calculate the magnetic field 10 cm away from a straight wire carrying 2 A of current.
Asked: magnetic field B
Given: distance r = 0.1 m; electric current I = 2 A
Relationships: B = μ0I/(2πr)
Solution: B= μ 0 I 2πr = (4π× 10 7  T m/A)(2 A) 2π(0.1 m) =4× 10 6  T
Answer: B = 4 × 10−6 T = 4 μT
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An average value of the Earth’s magnetic field is 5×10−5 T. How far away from a wire carrying 1 A of current does the magnitude of the wire’s magnetic field equal the Earth’s magnetic field? Show

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