Question

Find force per unit length at P

\[\begin{align}
  & A{{.10}^{-4}}m \\
 & B{{.10}^{-4}}N/m \\
 & C.3\times {{10}^{-4}}N/m \\
 & D.0.3N/m \\
\end{align}\]

Answer 1

Hint: Here, the dimensions of the rectangular wire and the current in the wire is given, then we can find the force experienced due to the wire. Here, we need to find the force per unit length, and then we can substitute the values.
Formula: $\vec F=I\vec l\times \vec B$

Complete answer:
We know that electricity and magnetism are interrelated. We know that a current carrying conductor can produce a magnetic field around itself. And similarly, a varying magnetic field induces a current in the wire which is kept in the near surrounding. We know that a current carrying conductor when placed in a magnetic field experiences a force.
The force $\vec F$ experienced by a current $I$ carrying wire of length $\vec l$ which is placed in a magnetic field $\vec B$ is given as $\vec F=I\vec l\times \vec B=Ilbsin\theta$, where $\theta$ is the angle made by the length vector to the magnetic field.
The direction of the force is given by the right hand rule, where the thumb, index and the middle finger are extended outwards, and are perpendicular to each other. If the thumb points towards the direction of current, the index finger points towards the direction of the magnetic field, then the middle finger points to the direction of the force.
Let us assume that the P is the midpoint of the given wire. If the current flows through the wire, then from the right hand thumb rule, we can say that the magnetic field is into the paper, inside the ring.
We also know that the magnetic field $B$ for a straight wire is given as, $B=\dfrac{\mu_{0}i}{4\pi r}$. Clearly, the magnetic field at P is only due to the current in the wires above and below the point P. As the magnetic field is parallel to the point P, we can say that it is $0$.
Thus, we have, $B_{P}=2\times \dfrac{\mu_{0}i}{4\pi r}$.
$\implies B_{P}=\dfrac{2\mu_{0}\times 5}{4\pi \times 5\times 10^{-2}}$
$\implies B_{P}=\dfrac{2\mu_{0}}{4\pi\times 10^{-2}}$
Then, the $\dfrac{F}{dl}=iB$
$\implies \dfrac{F}{dl}=5\times\dfrac{2\mu_{0}}{4\pi\times 10^{-2}}$
$\implies \dfrac{F}{dl}=10\times 10^{-5}$
$\therefore \dfrac{F}{dl}=10^{-4}N/m$

Thus, the correct option is \[B{{.10}^{-4}}N/m\] .

Note:
We know that Force is a vector quantity, which has both magnitude and direction. Since the force is a cross product of both the line vector and the magnetic field, we can clearly say that the force is perpendicular to both the line vector and the magnetic field.