Magnetism
A hollow tube is carrying an electric current along its length distributed uniformly over its surface. The magnetic field :
Sometimes we show an idealised magnetic field which is uniform in a given region and falls to zero abruptly. One such field is represented in figure. Using Ampere's law over the path PQRS, show that such a field is not possible.
We know, $$\displaystyle \int B \times dl = \mu_0 i$$. Theoritically $$B = 0$$ at A
If, a current is passed through the loop PQRS, then
$$B = \dfrac{\mu_0 i}{2 (\ell + b)}$$ will exist in its vicinity.
Now, As the $$\vec{B}$$ at A is zero. So there'll be no interaction
However practically this is not true. As a current carrying loop, irrespective of its near about position is always affected by an existing magnetic field.
A hollow tube is carrying an electric current along its length distributed uniformly over its surface. The magnetic field :
A steady electric current is flowing through a cylindrical conductor
A long straight wire of radius R carries a current distributed uniformly over its cross-section. The magnitude of the magnetic field is
In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero :
Amperes circuital law is given by
The coil of a dynamo is rotating in a magnetic field. The developed induced emf changes and the number of magnetic lines of force also changes. Which of the following conditions is correct?
State Ampere's circuital law.
A current of $$\dfrac{1}{(4\pi)}$$ ampere is flowing in a long straight conductor. The line integral of magnetic induction around a closed path enclosing the current-carrying conductor is
Figure shows an Amperian path $$ABCDA$$. Part $$ABC$$ is in verical plane $$PSTU$$ while part $$CDA$$ is in horizontal plane $$PQRS$$. Direction of circulation along the path is shown by an arrow near point $$B$$ and $$D$$. $$\oint { \vec { B } .d\vec { l } } $$ for this path according to Ampere's law will be :