Advertisements
Advertisements
प्रश्न
State Ampere’s circuital law.
State Ampere’s 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.
संबंधित प्रश्न
Write Maxwell's generalization of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is `I=varepsilon_0 (dphi_E)/dt,`where ΦE is the electric flux produced during charging of the capacitor plates.
Write Maxwell's generalization of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is `I=varepsilon_0 (dphi_E)/dt,`where ΦE is the electric flux produced during charging of the capacitor plates.
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?
Obtain an expression for magnetic induction along the axis of the toroid.
Explain 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.
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?
A long, straight wire carries a current. Is Ampere's law valid for a loop that does not enclose the wire, or that encloses the wire but is not circular?
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.
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.
A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross section. Find the magnetic field B at a point at a distance (a) 2 cm (b) 10 cm and (c) 20 cm away from the axis. Sketch a graph B versus x for 0 < x < 20 cm.
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.
Consider the situation of the previous problem. A particle having charge q and mass mis projected from the point Q in a direction going into the plane of the diagram. It is found to describe a circle of radius r between the two plates. Find the speed of the charged particle.
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).
What is magnetic permeability?
Define ampere.
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 ______.
Ampere’s circuital law is given by _______.
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:
A solenoid of length 0.6 m has a radius of 2 cm and is made up of 600 turns If it carries a current of 4 A, then the magnitude of the magnetic field inside the solenoid is:
A long solenoid having 200 turns per cm carries a current of 1.5 amp. At the centre of it is placed a coil of 100 turns of cross-sectional area 3.14 × 10−4 m2 having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within 0.05 sec, the induced e.m.f. in the coil is:
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:
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 ______.
A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field `vecB_1` at `a/2` and `vecB_2` at distance 2a is ______.
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.
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.
Briefly explain various ways to increase the strength of the magnetic field produced by a given solenoid.
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.
Which law provides a simpler alternative method for finding the magnetic field \[B\] when the current distribution has symmetry (straight wire, solenoid, toroid)?
The SI unit of electric current, the ampere (A), is named in honour of:
Substituting \[B=\frac{\mu_0 I}{2\pi r}\] into \[\oint\vec{B}\cdot d\vec{l}=B(2\pi r)\] for a straight wire yields:
In the sign convention for Ampere's law, if the fingers of the right hand curl along the direction of integration (\[d\vec{l}\]), the thumb points:
