Ampere's Circuital Law

Current I is following along the path ABCD, along the four edges of the cube (figure-a) creates a magnetic field in the centre of the cube of ${\text{B}}_0$. Find the magnetic field B created at the centre of the cube current I following along the path of the six edges ABCDHEA figure - b

Two long conductors are arranged as shown above to form overlapping cylinders, each of radius r, whose centres are separated by a distance d. Current of density J flows into the plane of the page along the shaded part of one conductor and an equal current flows out of the plane of the page along the shaded portion of the other, as shown. What are the magnitude and direction of the magnetic field at point A ?

A cylindrical conductor of radius $R$ carries a current along its length. The current density $J$, however, is not uniform over the cross-section of the conductor but is a function of the radius according to $J=br$, where $b$ is a constant. Then the expression for the magnetic field



$\left(B_1\right)$ at $r_1<R$ and
$\left(B_{\mathrm{2}}\right)$ at distance $r_2>R$, measured from the axis is:

Six wires of current ${\mathrm{I}}_1=\mathrm{1 }\mathrm{A},{\mathrm{I}}_2=\mathrm{2 }\mathrm{A},{\mathrm{I}}_3=\mathrm{3 }\mathrm{A},{\mathrm{I}}_4=\mathrm{1 }\mathrm{A},{\mathrm{I}}_5=\mathrm{5 }\mathrm{A}$ and ${\mathrm{I}}_6=\mathrm{4 }\mathrm{A}$ cut the page perpendicularly at the points $\mathrm{1,\; 2,\; 3,\; 4,\; 5}$ and $6$ respectively as shown in the figure. Find the value of integral $\oint \overrightarrow{B}.d\overrightarrow{l}$ around the circular path $\mathrm{C.}$

A cylindrical cavity of diameter$a$ exists inside a cylinder of diameter $2a$ shown in the figure, Both the cylinder and the cavity are infinitely long. A uniform current density$J$ flows along the length. If the magnitude of the magnetic field at the point $P$ is given by $\frac{N}{12}{\mu }_{0}aJ$ then the value of$N$ is: 

If a magnetic field exists in space $B = B_1\hat{i}+B_2\hat{j}$ then the plane through which magnetic flux will be zero is

A current carrying wire of length $l$ is first converted into a single loop ring and then into a $3$ loop coil. If magnetic field due to first ring is $B_1$, and due to second coil is $B_2$, at their respective centres, then the value of $\frac{B_1}{B_2}$ is

Two concentric circular coils with radii $1 \mathrm{cm}$ and $1000 \mathrm{cm}$ and number of turns $10$ and $200$ respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be _____ $\times {10}^{–8} \mathrm{H}$. (Take, ${\pi }^2=10$)

Two concentric and coplanar circular loops P and Q have their radii in the ratio $2:3$. Loop Q carries a current $9 \mathrm{A}$ in the anticlockwise direction. For the magnetic field to be zero at the common centre, loop P must carry

A closed curve encircles several conductors.The line integral $\int \overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{dl}}$ around this curve is $3.83\times {10}^{-7} \mathrm{T}-\mathrm{m}$. What is the net current in the conductors?

Two insulated wires of infinite length are lying mutually at right angles to each other as shown in. Currents of $2 \mathrm{A}$ and $1.5 \mathrm{A}$ respectively are flowing in them. The value of magnetic induction at point P will be-

Only the current inside the amperian loop contributes in

The figure shows the cross-section of two long co-axial tubes carrying equal current in opposite direction at point 1 and 2, as shown in the figure then

A closed curve encircles several conductors. The line integral $\int \overrightarrow{B}.d\overrightarrow{l}$ around this curve is $3.83\times {10}^{-7} \mathrm{T}·\mathrm{m}$. What is the net current in the conductors?

A closed curve encircles several conductors. The line integral$\int \overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{dl}}$  around this curve is $3.83\times {10}^{-7} \mathrm{T}-\mathrm{m}$. What is the net current in the conductors? 

Rank the magnitudes of $\oint \overrightarrow{B}·d\overrightarrow{l}$ for the closed paths in figure, from least to greatest.

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A toroid has a core (non-ferro magnetic) of inner radius $24 \mathrm{cm}$ and outer radius $26 \mathrm{cm}$ around which $2000$ turns of a wire is wound. If the current in the wire is $12 \mathrm{A}$, the magnetic field inside the core of the toroid is

A toroid has a non ferromagnetic core of inner radius $24 \mathrm{cm}$ and outer radius $25 \mathrm{cm}$, around which $4900$ turns of a wire are wound. If the current in the wire is $12 \mathrm{A}$, the magnetic field inside the core of the toroid is

Four wires carrying current ${\mathrm{I}}_1=2 \mathrm{A},{\mathrm{I}}_2=4 \mathrm{A},{\mathrm{I}}_3=6 \mathrm{A}$ and ${\mathrm{I}}_4=8 \mathrm{A}$ respectively cut the page perpendicularly as shown in figure. The value of $\int \overrightarrow{\mathrm{B}}·\mathrm{d}\overrightarrow{l}$ for the loop shown would be:

Consider six wires coming into or out of the page, all with the same current. Rank the line integral of the magnetic field (from most positive to most negative) taken counterclockwise around each loop shown.