हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

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. - Physics

Advertisements
Advertisements

प्रश्न

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.

टिप्पणी लिखिए
Advertisements

उत्तर

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Magnetic Field due to a Current - Exercises [पृष्ठ २५२]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 13 Magnetic Field due to a Current
Exercises | Q 53 | पृष्ठ २५२

संबंधित प्रश्न

State 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 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.


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.


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.


What is magnetic permeability?


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


The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?


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 ______.


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?


The force required to double the length of a steel wire of area 1 cm2, if it's Young's modulus Y = `2 xx 10^11/m^2` 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 ______


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.

  1. Explain the reason for the jumping of the ring when the switch is closed in the circuit.
  2. What will happen if the terminals of the battery are reversed and the switch is closed? Explain.
  3. Explain the two laws that help us understand this phenomenon.

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×