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Wize High School Grade 12 Physics Textbook > Magnetism

Magnetic Field of Current Carrying Wires

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Magnetic Field of Current-Carrying Wires


Current-carrying wires produce magnetic fields. In this section we'll see the formulas that connect the currents to the magnetic fields produced.
  • For a long, straight wire, the magnetic field experienced at a perpendicular distance rr from the wire is as follows:
B=μoI2πr\boxed{B=\frac{\mu_o I}{2\pi r}}
  • The direction is given by the right-hand rule. If you point your right thumb in the direction of the current, your fingers show whether the magnetic field is clockwise or counter-clockwise.
  • In this formula we have introduced the permeability of free space:
μo=4π×10−7T mA\boxed{\mu_o=4\pi\times10^{-7} \frac{T~m}{A}}

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Solenoids

  • A solenoid is a tightly wound coil of current.
  • The magnetic field outside a solenoid is zero.
  • The magnetic field inside a solenoid is given as follows:
B=μoNIL\boxed{B=\frac{\mu_o NI}{L}}
  • In this formula, L is the length of the solenoid, and N is the number of twists in the coil.
  • We can also re-write this formula in terms of current loops per unit length, n=NLn=\frac{N}{L}:
B=μonI\boxed{B=\mu_o nI}
  • The direction is given by the right-hand rule. If you curl your right-hand fingers along the direction of current, your thumb will show the direction of the magnetic field inside the solenoid.

Current Loop

  • At the center of a loop of current, the magnetic field is given by the following equation, where RR is the radius of the current loop:
B=μoI2R\boxed{B=\frac{\mu_o I}{2R}}
  • Once again, the magnetic field direction is found with the right-hand rule.

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Example: Magnetic Field Produced by a Wire


A current of 4.50 A flows through a long, straight conducting wire. A positive charge of 2.50 x 10-15 C travels at 1.24 x 103 m/s parallel to the wire, in the same direction of the current.

a) What is the magnitude of the field created by the wire at 5 cm from the wire?
b) Determine the magnitude and direction of the magnetic force exerted on the charge.

Part a)

We can find the magnitude of the field easily, by using the formula for the magnetic field strength at some distance from the wire:
∣B⃗∣=μoI2πr∣B⃗∣=(4π×10−7T mA)(4.5A)2π(0.05m)∣B⃗∣=1.8×10−5 T\begin{aligned} |\vec B|&=\frac{\mu_o I}{2\pi r}\\ |\vec B|&=\frac{(4\pi \times 10^{-7} \frac{T~m}{A}) (4.5 A)}{2\pi (0.05m)}\\ |\vec B|&=1.8\times10^{-5}~T \end{aligned}

Part b)

First let's find the magnitude of the force.
∣F⃗∣=qvBsin⁡θ∣F⃗∣=(2.5×10−15C)(1.24×103m/s)(1.8 T)sin⁡(90o)∣F⃗∣=55.8×10−18 N\begin{aligned} |\vec F|&=qvB\sin\theta \\ |\vec F|&=(2.5\times10^{-15}C)(1.24\times10^3 m/s)(1.8~T)\sin(90^o) \\ |\vec F|&=55.8\times10^{-18}~N \end{aligned}
The charge is travelling in the same direction as the current, and we know the field points either in or out of the page, which is how we found the angle to be 90 degrees.

Let's assume that the wire travels up the page and that the charge is on the right side of the wire (also travelling up the page). Our right-hand rule gives that the magnetic field from the wire must point in to the page.

For the direction of the force, we use the right-hand rule with the velocity and magnetic field. Since velocity points up the page and the field points into the page, we set our four fingers in the direction of the velocity (up) and curl them into the page. This shows a force going to the left, i.e. the force points towards the wire.

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Example: Magnetic Field from a Solenoid


What is the strength of the magnetic field in an infinitely long solenoid carrying 800 mA of current if for every 5 cm of the solenoid it has 200 loops?

This question is simply practicing with the formula for the magnetic field of a solenoid. Don't forget to convert all provided values to SI units.
B=μoNILB=(4π×10−7TmA)(200)(0.800A)(0.05m)B=4.02 mT\begin{aligned} B&=\cfrac{\mu_oNI}{L} \\ B&=\cfrac{(4\pi \times 10^{-7} \frac{Tm}{A})(200)(0.800 A)}{(0.05m)} \\ B &= 4.02 ~mT \end{aligned}

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Practice: Magnetic Field created by 2 Straight Wires

Consider two infinite wires that are parallel. They each carry current I and are separated by a distance L.

If the currents are in the same direction:
a) What is the magnetic field halfway between the wires?
b) What is the magnetic field a distance L away from the wires?

c), d): Repeat (a) and (b) if the currents are carried in opposite directions.

Practice: Magnetic Field of Current-Carrying Wires


Four straight wires, sitting at each corner of a square, are carrying equal electric currents I perpendicular to the page, as shown below.

a) Find the direction of net magnetic field at the center of square.
b) What is the expression for the magnitude of magnetic field at the center of the square?

Part a)
checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Practice: Magnetic Field of a Loop-de-Loop


Consider a very long wire carrying a current of I. The wire is twisted so that it forms a circular loop as shown below with a radius of R.

a) What is the magnetic field at the center of the loop?
b) Repeat part (a) if the wire is wound around the loop three times instead of just one.


Practice: Magnetic Field and Force on Straight Wires

A 25-cm long straight wire is on top of another wire which carries current of 2.0 A, as shown below. The top wire levitates motionless 5 mm above the bottom wire. What is the electric current of the top wire, and in what direction, if the top wire weighs 0.25 grams?
Magnitude of current