English
Karnataka Board PUCPUC Science Class 11

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.

Advertisements
Advertisements

Question

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.

Short/Brief Note
Advertisements

Solution

Given: 
Charge = q
Mass = m
Radius = r
We know that the radius described by a charged particle in a magnetic field is given by

`r = (mv) /(qB)`

Using Ampere circuital law 

`int B .dl = mu_0i`

`⇒ B. dl = mu _0 kdl ` 

`⇒ B  = mu_0 k`

`⇒ v =(Bqr)/m = (mu_0kqr)/m`

shaalaa.com
  Is there an error in this question or solution?
Chapter 35: Magnetic Field due to a Current - Exercises [Page 252]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 35 Magnetic Field due to a Current
Exercises | Q 53 | Page 252

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 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 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.


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. 


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.


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?


State Ampere’s circuital law.


Ampere’s circuital law states that ______.

Ampere’s circuital law is equivalent to ______.

A long solenoid has a radius a and number of turns per unit length n. If it carries a current i, then the magnetic field on its axis is directly proportional to ______.

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:


Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C ______.

  1. `oint B.dl = +- 2μ_0I`
  2. the value of `oint B.dl` is independent of sense of C.
  3. there may be a point on C where B and dl are perpendicular.
  4. B vanishes everywhere on C.

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 ______.


When current flowing through a solenoid decreases from 5A to 0 in 20 milliseconds, an emf of 500V is induced in it.

  1. What is this phenomenon called?
  2. Calculate coefficient of self-inductance of the solenoid.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×