Magnetic field due to a long straight wire & force between two parallel wires - Magnetism

Magnetic field due to a long straight wire & force between two parallel wires

Lessons

Notes:

In this lesson, we will learn:

  • Magnetic field due to a long straight wire
  • Magnetic field midway between two currents
  • Forces between two parallel wires

Notes:

An electric current produces a magnetic field
  • The magnetic field surrounding the electric current in a long straight wire is such that the field lines are circles with the wire at the center.
  • The field strength at a given point would be greater if the current flowing in the wire were greater; BI B \propto I
  • The filed strength would be less at points farther from the wire B1r B \propto \frac{1}{r} ;


  • BI \quad B \propto I
    B1rB1rB= \quad B \propto \frac{1}{r} \qquad B \propto \frac{1}{r} \quad \Rightarrow \quad B = μ02πIr \large \frac{\mu_{0}}{2 \pi} \frac{I}{r},


The value of the constant μ0\mu_{0} , which is called the permeability of free space, is μ0\mu_{0} = 4π \pi × 10-7 T.m/AT.m/A

Magnetic field due to a long straight wire & force between two parallel wires


Magnetic Field Midway Between Two Currents

Two parallel wires 10.0cm apart carry currents in opposite directions. Current I1I_{1} = 5.0A is out of the page, I2I_{2}=7.0 A is into the page. Determine the magnitude and direction of the magnetic field halfway between the two wires.

Magnetic field due to a long straight wire & force between two parallel wires


B1=B_{1} = μ0I12πr=(4π×107T.m/A)(5.0A)2π(0.050m) \large \frac{\mu_{0} I_{1}} {2\pi r} = \frac{(4\pi \, \times \, 10^{-7} \, T\, . \, m/A) (5. 0A) } {2 \pi (0.050m) } = 2.0 × 10-5 TT

B2=B_{2} = μ0I22πr=(4π×107T.m/A)(7.0A)2π(0.050m) \large \frac{\mu_{0} I_{2}} {2\pi r} = \frac{(4\pi \, \times \, 10^{-7} \, T\, . \, m/A) (7. 0A) } {2 \pi (0.050m) } = 2.8 × 10-5 TT

The total filed is up with the magnitude of

B=B1+B2=4.8×105TB = B_{1} +B_{2} = 4.8 \times 10^{-5} T


Forces Between Two Parallel Wires
  • Consider two long parallel wires separated by a distance dd. They carry currents I1I_{1} and I2I_{2}, respectively. Each current produces a magnetic field that is felt by the other, so each must exert a force on the other.


  • Magnetic field due to a long straight wire & force between two parallel wires


  • Magnetic field B1B_{1} produced by I1,B1=I_{1}, \, B_{1} = μ0I12πd \large \frac{\mu_{0} I_{1}} {2 \pi d}
  • The force F2F_{2} is exerted by B1B_{1} on a length of I2I_{2} of wire 2, carrying current I2I_{2}, F2F_{2} = I2I_{2} B1B_{1} I2I_{2}
  • Substitute B1B_{1} into F2F_{2} formula to get the final equation;


  • F2=F_{2} = μ02πI1I2dI2 \large \frac{\mu_{0}} {2 \pi} \frac{I_{1} I_{2}} {d} I_{2}

  • Parallel currents in the same direction exert an attractive force on each other
  • Antiparallel currents (in opposite directions) exert a repulsive force on each other.

Magnetic field due to a long straight wire & force between two parallel wires
  • Intro Lesson
Teacher pug

Magnetic field due to a long straight wire & force between two parallel wires

Don't just watch, practice makes perfect.

We have over NaN practice questions in Physics for you to master.