Wire carrying current $I$ immersed in magnetic field $\vec B$ . $\vec F$ shows the force on the wire due to the $B$

• Recall $\vec F_B = q \vec \times \vec B$ and $d \vec R = dq \vec V \times \vec B$

  • But, $I = \frac{dq}{dt} \to dq = Idt$ and $V \frac{d \vec L}{dt}$, where $\vec L$ is the length vector

• $d \vec F = I dt \frac{d \vec L}{dt} \times \vec B = I d \vec L \times \vec B$

• To calculate the total force, we integrate $\int d \vec F = \int I d \vec L \times \vec B$

  • if we assume $F$ is constant, the wire’s current is straight and the length $L$ and $\vec B$ does not depend on $L$, then $\vec F = I \vec L \times \vec B$

    

• The force experiences by the wire can be understood in terms of charges.

  • If we assume current $I$ is due to negative charges, then $\vec F$ points to the right
  • If we assume current is due to positive charges, force $\vec F$ still points to the right
  • This force is then transferred to the body of the wire which results in wire bending to the right

  

Force bewteen two long wires carrying currents (paper)

We observe that the two wires attract to eachother. If the direction of current in one wire is changes, wires repell each other. How can we understand and explain this?

• Current $I_a$ in wire $a$ will produce a magnetic field $\vec B _a$ at the site of wire-b
• Magnitude of this field is given byb $B_a = \frac{\mu _0 I_a}{2 \pi d}$
  • Wire b is immersed in magnetic field $B_a$ (produced by wire a), $\vec F_{b/a} = \frac{\mu _0 L I_a I-b}{2 \pi d}$