English
Karnataka Board PUCPUC Science Class 11

A Hollow Tube is Carrying an Electric Current Along Its Length Distributed Uniformly Over Its Surface. the Magnetic Field

Advertisements
Advertisements

Question

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.

Short/Brief Note
Advertisements

Solution

(b) is constant inside the tube
(c) is zero at the axis

A hollow tube is carrying uniform electric current along its length, so the current enclosed inside the tube is zero.
According to Ampere's law, 

\[\oint \vec{B} . d \vec{l} = \mu_o i_{\text{inside}} \]
\[\text{ Inside the tube }, \]
\[\oint \vec{B} . d \vec{l} = 0, r < R\]
\[ \Rightarrow B_{\text{inside}} = \text{ Constant}\]
\[ \Rightarrow B_{\text{axis}} = 0 \]

The  magnetic fields from points on the circular surface will point in opposite directions and cancel each other.

Outside the tube, 
\[B \times 2\pi r = \mu_o i\]
\[ \Rightarrow B_{\text{outside}} = \frac{\mu_o i}{2\pi r}, r > R\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 35: Magnetic Field due to a Current - MCQ [Page 249]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 35 Magnetic Field due to a Current
MCQ | Q 5 | Page 249

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.


Explain Ampere’s circuital law.


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 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? 


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. 


What is magnetic permeability?


Define ampere.


Find the magnetic field due to a long straight conductor using Ampere’s circuital law.


Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law


Ampere’s circuital law is equivalent to ______.

The magnetic field around a long straight current carrying wire 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?


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:


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 ______.


Briefly explain various ways to increase the strength of the magnetic field produced by a given solenoid.


The SI unit of electric current, the ampere (A), is named in honour of:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×