Advertisements
Advertisements
Question
In Ampere's \[\oint \vec{B} \cdot d \vec{l} = \mu_0 i,\] the current outside the curve is not included on the right hand side. Does it mean that the magnetic field B calculated by using Ampere's law, gives the contribution of only the currents crossing the area bounded by the curve?
Advertisements
Solution
In Ampere's law \[\oint \vec{B} . \vec{dl} = \mu_o i\] , i is the total current crossing the area bounded by the closed curve. The magnetic field B on the left-hand side is the resultant field due to all existing currents.
APPEARS IN
RELATED QUESTIONS
State Ampere’s circuital law.
Electron drift speed is estimated to be of the order of mm s−1. Yet large current of the order of few amperes can be set up in the wire. Explain briefly.
A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
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.
A hollow tube is carrying an electric current along its length distributed uniformly over its surface. The magnetic field
(a) increases linearly from the axis to the surface
(b) is constant inside the tube
(c) is zero at the axis
(d) is zero just outside the tube.
In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero
(a) outside the cable
(b) inside the inner conductor
(c) inside the outer conductor
(d) in between the tow conductors.
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 ______.
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?
Ampere's circuital law is used to find out ______
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:
Two concentric and coplanar circular loops P and Q have their radii in the ratio 2:3. Loop Q carries a current 9 A in the anticlockwise direction. For the magnetic field to be zero at the common centre, loop P must carry ______.
Read the following paragraph and answer the questions.
|
Consider the experimental set-up shown in the figure. This jumping ring experiment is an outstanding demonstration of some simple laws of Physics. A conducting non-magnetic ring is placed over the vertical core of a solenoid. When current is passed through the solenoid, the ring is thrown off. |
- Explain the reason for the jumping of the ring when the switch is closed in the circuit.
- What will happen if the terminals of the battery are reversed and the switch is closed? Explain.
- Explain the two laws that help us understand this phenomenon.
Using Ampere’s circuital law, obtain an expression for magnetic flux density ‘B’ at a point near an infinitely long and straight conductor, carrying a current I.
