Advertisements
Advertisements
Question
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.
Advertisements
Solution
(a) The magnetic field inside any conducting tube is always zero.
∴ Magnetic field inside the tube at a distance r/2 from the surface = 0
(b) Let the point outside the tube with distance \[\frac{r}{2}\] be P.
∴ Net distance from centre, r' = \[r + \frac{r}{2} = \frac{3r}{2}\]
Length of the loop, l = \[2\pi \times \frac{3}{2}r = 3\pi r\]
On applying Ampere's law, we get
\[ \Rightarrow B \times 3\pi r = \mu_0 i\]
\[ \Rightarrow B = \frac{\mu_0 i}{3\pi r}\]
APPEARS IN
RELATED QUESTIONS
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 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?
In order to have a current in a long wire, it should be connected to a battery or some such device. Can we obtain the magnetic due to a straight, long wire by using Ampere's law without mentioning this other part of the circuit?
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.
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.
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.
State Ampere’s circuital law.
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 ______.
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:
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 ______.
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.
Magnetic fields are produced by:
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:
According to the right-hand rule sign convention for Ampere's law, currents flowing in the direction of the thumb are taken as ______ and currents flowing opposite to it are taken as ______.
