Advertisements
Advertisements
प्रश्न
State Ampere’s circuital law.
Advertisements
उत्तर
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
संबंधित प्रश्न
State Ampere’s circuital law.
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 ?
A long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. The current is uniformly distributed across the cross-section. Apply Ampere’s circuital law to calculate the magnetic field at a point ‘r’ in the region for (i) r < a and (ii) r > a.
Consider the situation described in the previous problem. Suppose the current i enters the loop at the points A and leaves it at the point B. Find the magnetic field at the centre of the loop.
A thin but long, hollow, cylindrical tube of radius r carries i along its length. Find the magnitude of the magnetic field at a distance r/2 from the surface (a) inside the tube (b) outside the tube.
A long, cylindrical tube of inner and outer radii a and b carries a current i distributed uniformly over its cross section. Find the magnitude of the magnitude filed at a point (a) just inside the tube (b) just outside the tube.
Two large metal sheets carry currents as shown in figure. The current through a strip of width dl is Kdl where K is a constant. Find the magnetic field at the points P, Q and R.
Calculate the magnetic field inside and outside of the long solenoid 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 ______.
In a capillary tube, the water rises by 1.2 mm. The height of water that will rise in another capillary tube having half the radius of the first is:
