Advertisements
Advertisements
Question
State Ampere’s circuital law.
Solution
The line integral of the magnetic field over a closed-loop is μ0 times the net current enclosed by loop.
`oint_"C" vec"B" * vec"l" = mu_circ "I"_"enclosed"`
`"I"_"enclosed"` → net current in the closed-loop C.
APPEARS IN
RELATED QUESTIONS
Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis ?
Using Ampere's circuital law, obtain an expression for the magnetic flux density 'B' at a point 'X' at a perpendicular distance 'r' from a long current-carrying conductor.
(Statement of the law is not required).
Define ampere.
Find the magnetic field due to a long straight conductor using Ampere’s circuital law.
A straight wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter carrying the same current. The strength of the magnetic field far away is ______.
Two identical current carrying coaxial loops, carry current I in opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C, then which statement is correct?
A thick current carrying cable of radius ‘R’ carries current ‘I’ uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance ‘r’ from the axis of the cable is represented by ______
The given figure shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a.
When current flowing through a solenoid decreases from 5A to 0 in 20 milliseconds, an emf of 500V is induced in it.
- What is this phenomenon called?
- Calculate coefficient of self-inductance of the solenoid.
