In the figure an electron with an initial kinetic energy of...

In the figure, an electron with an initial kinetic energy of 4.10 keV enters region 1 at time t = 0. That region contains a uniform magnetic field directed into the page, with magnitude 0.0120 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 30.0 cm. There is an electric potential difference ΔV = 2000 V across the gap, with a polarity such that the electron's speed increases uniformly as it traverses the gap. Region 2 contains a uniform magnetic field directed out of the page, with magnitude 0.0209 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave?

Transcript

0:00 Hi.

00:01 Here in this given problem there are two regions having different magnetic fields, uniform magnetic fields but opposite to each other.

00:17 This is b1 directed into the plane of paper in this rectangular region and then another magnetic field coming out of the plane of paper b2 and between these two regions there is no magnetic field an electron is said to be moving is said to be entering into this magnetic field perpendicularly then it will undergo a semi -circular path within this magnetic field it is said to be coming out, then it is moving straight towards the another magnetic field under the influence of an electric field due to which its energy will be increased.

01:08 And now it will enter into the another magnetic field and we'll start moving again in a circular path, we'll cover the semicircular path and will come out into the air again like this.

01:24 And we have to find the total time taken by this electron from this region when it enters into magnetic field b1 to when it comes out of the magnetic field b2.

01:37 The gap between these two magnetic fields, if we consider it to be delta x, it is given as 30 centimeter.

01:50 Means this will be 0 .30 meter.

01:56 Magnitude of magnetic field b1 this is 0 .0 .021 tesla.

02:04 Magnitude of magnetic field b2 this is 0 .029 tesla.

02:13 Potential difference between these two magnetic fields applied to accelerate the charge particle that is 2000 volt.

02:22 Now, first of all, we find time taken by the electron in this magnetic field, let it be t1.

02:30 Here, suppose it will be t2 and then between these two magnet fields, suppose it will be t3.

02:39 So first of all, time taken to cover a semi -circular trajectory in a magnetic field, the time will be given by an expression pi m by bq.

02:50 For q this is e.

02:52 So t1 will be pi m by b1e and t2 will be pi m by b2e.

03:01 So first of all we find this total time t1 plus t2 taking this pi m by e as a common out 1 by b1 plus 1 by b2 plugging in now all the known values this is 3 .14 into mass of electron 9 .1 into 10 dash per minus 31 divided by charge over electron 1 .6 into 10 dash the power minus 19 goulon.

03:34 1 by b1 which is 0 .0 120 tesla plus 1 by t2 sorry b2 which is 0 .0209 and and finally, it is solved to get 2 .34 into 10 dash the power minus 9 second, or we can say this is 2 .34 nanosecond, the two times.

04:05 Now we have to find time taken to cover the gap between two magnetic fields for which for the gap between the two fields, first of all, work done on this electron, it is given either by the product of charge with the potential difference through which it has been accelerated or simply by the product of force experienced by the electron with the displacement covered by it.

04:50 And force here may be given as ma, simply using newton second of the electron.

04:55 Motion into delta x.

04:57 So we get an expression for the acceleration in the motion of this electron and that is e into delta v divided by m delta x plugging in all known values.

05:10 This is 1 .6 into 10 dash to power minus 19 coulum for the charge over electron 2 ,000 volt for the potential difference mass 9 .1 into 10 dash 1 minus 31 into delta x 0 .30.

05:25 So, acceleration here comes out to be equal to 1 .17 into 10 -dish to the power 15 meter per second square.

05:36 Now we will find its speed also as it will be coming out of the field here...

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