An electron moving in a circular orbit of radius R completes n round per second find the magnitude of the magnetic field at the centre?
An electron moving in a circular orbit of radius R completes n rounds per second find the magnitude of the magnetic field at the centre?The electron completes ‘n’ rounds per second, which meansfrequency = nTherefore, [math]T = \dfrac{1}{n} (As T = \dfrac{1}{f})[/math]As [math]I = \dfrac{q}{T}[/math]Charge of an electron; [math]q=e (e= 1.6×10^{-19})[/math][math]I = \dfrac{e}{1/n}[/math]i.e, I = n×eMagnetic field at a point P on the central axis of a circular loop which is at a distance ‘x' from the centre of the circular loop of radius 'R' is given by[math]B = \dfrac{μ0×I×r^{2}}{2×(x^{2}+r^{2})^{\dfrac{3}{2}}}[/math]As x=0 (because the question asks for magnetic field at the center of the loop)[math]B = \dfrac{μ0×I}{2×r}[/math]Substitute I=n×ewe get;[math] B = \dfrac{μ0×n×e}{2×r}[/math]
An electron has a circular path of radius 0.01m in a perpendicular magnetic induction 0.001 Tesla. What is the speed of the electron?
Any objects which moves in a circular motion around a centre of any thing is under centripetal force and this force is responsible for the circular motion motion of the object and its magnitude isForce= mass×velocity square / radiusNow as electron is moving in a circular path , so centripetal force is responsible for its circular motion and also you know that when electron moves in a magnetic field , its experienced a force = q v ×B , due to this force electron is moving in a circular path .So any force or anything which responsible for circular motion is same as the centripetal force . So , magnitude of force experienced by electron in a magnetic field is same as centripetal force acting on the electron .Centripetal force = force experienced by elctron in a magnetic field .mv2/r=qvB , put the values and find the value of speed of electron
Electron and proton are moving with same speed.Which will have more wavelength?
More is the mass, more will it interact with the Higgs Field and due to The Greater Interaction, Less Movement would be possible. If Less Movement is possible then Shorter Wavelengths would be assigned. That is why a Mathematical Formula (Given by De-Broglie) came up, where wavelength is inversely propotional to the magnitude of mass.Conclusively, we can say, that, as Proton has more mass, it will have a Shorter wavelength than an Electron moving with the same speed.Hope it Helps! Enjoy :D
If the charge on an electron is 1.6×10 ^(-19) C, find the approximate number of electrons in 1C?
Just find the reciprocal of 1.6*10(-19) and you will have the result you are requesting. (6.25 *10^18) Check it.
A body moving with an initial velocity of 20 m/s has an acceleration of 4 m/s^2. Find the velocity of the moving body after 4s and distance traveled in that time?
In order to answer this question, we must apply newton's first equation of motionv = u + atWhere v is the final velocity in m/s, which is we shall attempt to find,u is the initial velocity in m/s, given in the question as 20m/sa is the acceleration of the body in m/s2, in the question, it is indicated to be 4m/s2t is the time in seconds, from the question, it is 4s.v = 20 + 4*4v = 20 + 16 = 36m/sAs for the distance, either the second or third equation of motion can be used.Using the second equation,s (distance) = u*t + 0.5*a*(t^2)s = 20*4 + 0.5*4*(4*4)s = 80 + 32 = 112mUsing the third equation of motion,v2 = u2 + 2as36^2 = 20^2 + 2*4*s1296 = 400 + 8 * s8*s = 1296 - 400 = 896s = 896/8 = 112m
If an electron is accelerated through a potential difference of 45.5 volts, then what is the velocity acquired by it?
Whenever a charged particle q is accelerated through a potential difference of V, its kinetic energy increases by an amount of qV. This you can deduce simply by applying the total mechanical energy conservation principle on the system as follows:Now in the given problem q = e = charge of an electron, m = mass of an electron and potential difference V = 45.5 volts. On putting these values in the above expression we getv = 4 × 10^6 m/s.Regards!
A beam of electrons passes undeviated through a region, what can be said about existence of electric and magnetic field?
If a beam of electrons passes through a region undeviated then there exists both magnetic and electric field perpendicular to each other. The velocity of electron beam would be E/B where E is the electric field and B is the magnetic field.
What is drift velocity and what is the relation between current and drift velocity?
The motions of free electrons( which are charge carriers) in conductors, are quite different from their motions in vacuum.When we apply voltage across a conductor, immediately an electric field is set up in the entire conductor.Let us look at motion of a single electron.The electron acquires an accelerated motion under the influence of electric field in vacuum, but within the conductor it moves with certain average velocity.Let us understand as to how this happens.We know that a conductor ( metal) contains free electrons and vibrating ions at the lattice sites. The free electrons move on zigzag paths. One such path for an electron is shown in figure (1) below with a continuous zig zag path.When electron moves, it actually moves in the continuously and randomly changing electric fields of the vibrating ions. When an electron comes in the vicinity of an ion, the effect of it's changing field is maximum and path of the electron changes abruptly as if it has collided with the ion.Now, suppose, applied voltage produces an electric field E. This field exerts a force -eE on the electron in the direction opposite to field. This is additional force over and above the abrupt force applied by ions as stated above.The electric field accelerates the electrons while ions scatter the electrons and change the directions of motions. Thus, we have zigzag paths of electrons. In figure (1) full line is path without electric field while dashed line shows path under the effect of field plus ions.In the absence of field electron moves from A to B. But, in the presence of voltage( hence field) it moves from A to B' as shown in the fig. It is as if the electron has been dragged through effective distance BB'. The velocity corresponding to this BB' distance is known as drift velocity denoted by v(d).Now, look at figure (2). Let N be the number of electrons passing through cross section per unit time in the direction normal to cross section.Therefore ,current by definition, isI=Ne. e is electronic charge.All the electrons passing through a cross section cover a distance v(d) in 1s and are contained in volume v(d)x A x 1 = A x v(d) as shown in fig.(2). A is area of cross section.If n is electron number density, N = n A v(d) orI =n A e v(d).Let us now calculate v(d). If average time interval between two consecutive ' collisions' with ions is T, the average velocity will bev(d)=a T= (Ee/m) T.T is called relaxation time.
The potential difference between 2 plates separated by a distance of 1 mm is 100 v. The force on an electron placed in between the plates is what?
Electric field E = V/d, where V is potential difference, and d is distance between the two plates. so, here E= 100 V/1mm = 100/(10^-3)=100x10^3=10^5 (Volt/m)So force on a electron in electric field is F= E.q = (10^5)x(1.6x10^-19) Newton