Question
State Ampere’s law. Explain how is it useful in different situations.
✓
Answer
Ampere’s circuital law : In free space, the line integral of magnetic induction around a closed path in a magnetic field is equal to p0 times the net steady current enclosed by the path.
In mathematical form,$\oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 \mid$
where $\vec{B}$ is the magnetic induction at any point on the path in vacuum, $\overrightarrow{d l}$ is the length element of the path, $I$ is the net steady current enclosed and $\mu_0$ is the permeability of free space.
Explanation : Figure shows two wires carrying currents $I_1$ and $I_2$ in vacuum. The magnetic induction $\vec{B}$ at any point is the net effect of these currents.
To find the magnitude $B$ of the magnetic induction :
We construct an imaginary closed curve around the conductors, called an Amperian loop, and imagine it divided into small elements of length $\overrightarrow{d l}$. The direction of dl is the direction along which the loop is traced.
(ii) We assign signs to the currents using the right hand rule : If the fingers of the right hand are curled in the direction in which the loop is traced, then a current in the direction of the outstretched thumb is taken to be positive while a current in the opposite direction is taken to be negative.
For each length element of the Amperian loop, $\vec{B} \cdot \overrightarrow{d l}$ gives the product of the length $dl$ of the element and the component of $\vec{B}$ parallel to $\overrightarrow{d l}$. If $\theta r$ is the angle between $\overrightarrow{d l}$ and $\vec{B}$
$\vec{B} \cdot \overrightarrow{d l}=(B \cos \theta) d l$
Then, the line integral,
$\oint \vec{B} \cdot \overrightarrow{d l}=\oint B \cos \theta d l$
For the case shown in Fig., the net current I through the surface bounded by the loop is
$ I = I _2- I _1$
$\therefore \oint B \cos \theta d l=\mu_0 I$
$=\mu_0\left( I _2- I _1\right) \ldots \ldots \ldots \ldots . \text { (3) } $
Equation (3) can be solved only when B is uniform and hence can be taken out of the integral.
[Note: Ampere's law in magnetostatics plays the part of Gauss's law of electrostatics. In particular, for currents with appropriate symmetry, Ampere's law in integral form offers an efficient way of calculating the magnetic field. Like Gauss's law, Ampere's law is always true (for steady currents), but it is useful only when the symmetry of the problem enables B to be taken out of the integral $\oint \vec{B} \cdot \overrightarrow{d l}$. The current configurations that can be handled by Ampere's law are infinite straight conductor, infinite plane, infinite solenoid and toroid.]
In mathematical form,$\oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 \mid$
where $\vec{B}$ is the magnetic induction at any point on the path in vacuum, $\overrightarrow{d l}$ is the length element of the path, $I$ is the net steady current enclosed and $\mu_0$ is the permeability of free space.
Explanation : Figure shows two wires carrying currents $I_1$ and $I_2$ in vacuum. The magnetic induction $\vec{B}$ at any point is the net effect of these currents.
To find the magnitude $B$ of the magnetic induction :
We construct an imaginary closed curve around the conductors, called an Amperian loop, and imagine it divided into small elements of length $\overrightarrow{d l}$. The direction of dl is the direction along which the loop is traced.
(ii) We assign signs to the currents using the right hand rule : If the fingers of the right hand are curled in the direction in which the loop is traced, then a current in the direction of the outstretched thumb is taken to be positive while a current in the opposite direction is taken to be negative.
For each length element of the Amperian loop, $\vec{B} \cdot \overrightarrow{d l}$ gives the product of the length $dl$ of the element and the component of $\vec{B}$ parallel to $\overrightarrow{d l}$. If $\theta r$ is the angle between $\overrightarrow{d l}$ and $\vec{B}$
$\vec{B} \cdot \overrightarrow{d l}=(B \cos \theta) d l$
Then, the line integral,
$\oint \vec{B} \cdot \overrightarrow{d l}=\oint B \cos \theta d l$
For the case shown in Fig., the net current I through the surface bounded by the loop is
$ I = I _2- I _1$
$\therefore \oint B \cos \theta d l=\mu_0 I$
$=\mu_0\left( I _2- I _1\right) \ldots \ldots \ldots \ldots . \text { (3) } $
Equation (3) can be solved only when B is uniform and hence can be taken out of the integral.
[Note: Ampere's law in magnetostatics plays the part of Gauss's law of electrostatics. In particular, for currents with appropriate symmetry, Ampere's law in integral form offers an efficient way of calculating the magnetic field. Like Gauss's law, Ampere's law is always true (for steady currents), but it is useful only when the symmetry of the problem enables B to be taken out of the integral $\oint \vec{B} \cdot \overrightarrow{d l}$. The current configurations that can be handled by Ampere's law are infinite straight conductor, infinite plane, infinite solenoid and toroid.]
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In the experiment to investigate the phenomenon of electromagnetic induction for a magnet swinging through a coil, show that the peak induced emf is directly proportional to the speed of the magnet $($or show that the peak induced emf is directly proportional to the angular amplitude and inversely proportional to the time period$).$
→State and prove Kirchoff’s law of heat radiation.
What is an electromagnet?
A circular coil, having $200$ turns each of area $2.5 \times 10^{-4} m ^2$, carries a current of $200 \mu A$. Initially, the coil is at rest in a magnetic field of induction $0.8 T$, with its magnetic dipole moment aligned with the field. Find the work an external agent has to do to rotate the coil through $(i)\ 90^{\circ}$ from its initial position $(ii)$ further $90^{\circ}$.
→Alternating emf of e = 220 sin 100 πt is applied to a circuit containing an inductance of (1/π) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
In an adiabatic compression of a gas the final volume of the gas is $80 \%$ of the initial volume. If the initial temperature of the gas is $27^{\circ} C$, find the final temperature of the gas. Take $\gamma=$ $5 / 3$.
→A cycle wheel with $10$ spokes, each of length $0.5 m$, is moved at a speed of $18 \ km / h$ in a plane normal to the Earth's magnetic induction of $3.6 \times 10^{-5} T$. Calculate the emf induced between
$(i)$ the axle and the rim of the cycle wheel
$(ii)$ ends of a single spoke and ten spokes.
→$(i)$ the axle and the rim of the cycle wheel
$(ii)$ ends of a single spoke and ten spokes.
A blackbody at $327^{\circ} \mathrm{C}$, when suspended in a black enclosure at $27^{\circ} \mathrm{C}$, loses heat at a certain rate. Find the temperature of the body at which its rate of loss of heat by radiation will be half of the above rate. Assume that the other conditions remain unchanged.
→Explain the Rayleigh criterion for the limit of resolution for
(i) two linear objects
(ii) a pair of point objects.
(i) two linear objects
(ii) a pair of point objects.
State the laws of simple pendulum.