If a current carrying wire is placed inside a magnetic field(B), the wire will produce it's own magnetic field that will oppose the applied external magnetic field(B). Due to the two fields opposing one another, would that somehow change the ϕ acting on the wire inducing an ϵ to oppose that change?

I'm aware that this wire would experience the Lorentz Force, and from that will move and from that motion, there is induced ϵ = −vBL.

  • 1
    \$\begingroup\$ Why would a current carrying wire in a magnetic field produce its own magnetic field in opposition? Isn't the magnetic field around a current-carrying wire a function of the current in that wire, and wouldn't whether it opposes some other field or not depend on what that other field is? \$\endgroup\$
    – Phil Frost
    Feb 16, 2015 at 20:48
  • \$\begingroup\$ @PhilFrost I'm not sure myself... But the wire would produce a magnetic field, I related Lenz-law to assume it's in opposition. I'm not sure how that wire's field would interact with the exterior field, could you clarify my confusion here? \$\endgroup\$
    – Pupil
    Feb 16, 2015 at 21:43

1 Answer 1


A current carrying wire will always generate its magnetic field, if this is going to be opposite to the external field or not, depends in which direction the current flows.

That beeing said, there will be no induced voltage if there will be no flux change. For flux to change you can either keep the outer field constant and change the area (e.g. move the wire) or you keep the area constant and change the external field, or change the current in the wire, thus changing the total flux in the area.

  • \$\begingroup\$ Wouldn't the wire's own magnetic field oppose/attract the external magnetic field causing a flux change? \$\endgroup\$
    – Pupil
    Feb 18, 2015 at 4:26
  • \$\begingroup\$ Assuming the wire's current is constant. \$\endgroup\$
    – Pupil
    Feb 18, 2015 at 5:12
  • \$\begingroup\$ Or to be exact, the wire's magnetic field would increase/decrease the total magnetic field( B(External)+ B(wire) = Btot)causing change in magnetic flux, and a induced-EMF to oppose that? \$\endgroup\$
    – Pupil
    Feb 18, 2015 at 7:20
  • \$\begingroup\$ Yes. However once the flux is static no voltage will be induced. So the way you described the problem, the change already happened and you have a static flux, so no induced voltage. \$\endgroup\$
    – iggy
    Feb 18, 2015 at 22:32
  • \$\begingroup\$ In a motor like application when might you think this could happen? Or when a current-carrying wire is placed in a magnetic field and starts to move due to the Lorentz force? \$\endgroup\$
    – Pupil
    Feb 18, 2015 at 22:36

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