Question 12 Marks
$a$. State Ampere's circuital law connecting the line integral of $B$ over a closed path to the net current crossing the area bounded by the path.
$b$. Use Ampere's law to derive the formula for the magnetic field due to an infinitely long straight current carrying wire.
$c$. Explain carefully why the derivation as in $(b)$ is not valid for magnetic field in a plane normal to a currentcarrying straight wire of finite length and passing through the midpoint of the axis.
$b$. Use Ampere's law to derive the formula for the magnetic field due to an infinitely long straight current carrying wire.
$c$. Explain carefully why the derivation as in $(b)$ is not valid for magnetic field in a plane normal to a currentcarrying straight wire of finite length and passing through the midpoint of the axis.
Answer
View full question & answer→$a$. Ampere's Circuital Law: It gives the relationship between the current and the magnetic field created by it.
This law says that the integral of magnetic field density $(B)$ along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.
$b.$ Now, from Ampere's Circuital law, $\int B . dI =\mu_0 l$
$\Rightarrow \int B \cdot dl =\mu_0 l \ [$Since angle between $B$ and $dl$ is zero$]$
$\Rightarrow B \int dl =\mu_0 l$
$\Rightarrow B .2 \pi r =\mu_0 l\ [$Since the total length of dl is equal to circumference of the circle considered$]$
$\Rightarrow B =\frac{\mu_0 l}{2 \pi r}$ which is the required expression for magnetic field of induction due to a long straight current carrying wire.
$c$. A straight conductor of finite length cannot by itself form a complete, steady current circuit.
Additional conductors are necessary to close the circuit.
These will spoil the symmetry of the problem.
The difficulty disappears if the conductor is infinitely long.
This law says that the integral of magnetic field density $(B)$ along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.
$b.$ Now, from Ampere's Circuital law, $\int B . dI =\mu_0 l$
$\Rightarrow \int B \cdot dl =\mu_0 l \ [$Since angle between $B$ and $dl$ is zero$]$
$\Rightarrow B \int dl =\mu_0 l$
$\Rightarrow B .2 \pi r =\mu_0 l\ [$Since the total length of dl is equal to circumference of the circle considered$]$
$\Rightarrow B =\frac{\mu_0 l}{2 \pi r}$ which is the required expression for magnetic field of induction due to a long straight current carrying wire.
$c$. A straight conductor of finite length cannot by itself form a complete, steady current circuit.
Additional conductors are necessary to close the circuit.
These will spoil the symmetry of the problem.
The difficulty disappears if the conductor is infinitely long.
