A particle of mass `m=1.6xx10^-27` kg and charge `q=1.6xx10^-19C` enters a region of uniform magnetic field of stregth `1 T` along the direction shown in figure. The speed of the particle is `10^7 m//s`
a. The magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the fiedl at the point `F`. Find the distasnce `EF` and the angle theta.
b. If the direction of the field is along the outward normal to the plane of the paper find the time spent by the particle in the regin of the magnetic field after entering it at `E`.
A particle of mass `m=1.6xx10^-27` kg and charge `q=1.6xx10^-19C` enters a region of uniform magnetic field of stregth `1 T` along the direction shown in figure. The speed of the particle is `10^7 m//s`
a. The magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the fiedl at the point `F`. Find the distasnce `EF` and the angle theta.
b. If the direction of the field is along the outward normal to the plane of the paper find the time spent by the particle in the regin of the magnetic field after entering it at `E`.
a. The magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the fiedl at the point `F`. Find the distasnce `EF` and the angle theta.
b. If the direction of the field is along the outward normal to the plane of the paper find the time spent by the particle in the regin of the magnetic field after entering it at `E`.
Text Solution
Verified by Experts
Inside a magnetic field, speed of charged particle does not change. Futher, velocity is perpendicular to magnetic field in both the cses hene path of the particle in the magnetic field will be circular. Centre of circle can be obtained by drawing perpendiculars to velocity (or tantent to the circular path) at `E` and `F`. Radius and angular speed of circular path would be
`r=(mv)/(Bq)` and `omega=(Bq)/m`
a. Refer figure i
`/_CFG=90^@-theta` and `/_CEG=90^@-45^@=45^@`
since `CF=CE`
`:. /_CFG=/_CEG`
or `90^@-theta=45^@` or `theta=45^@`
Further `FG=GE=rcos45^@`
`:. EF=2FG=2rcos45^0=(2mvcos45^@)/(Bq)`
`=(2(1.6xx10^-27)(10^7)(1/sqrt2))/((1)(1.6xx10^-19))=0.14m`
b. refer figure ii In this casse particle will complete `3/4` th of circle in the magnetic field.
Hence, the time spend in the magnetic field
`t=3/4` (time period of circular motion)
`=3/4((2pim)/(Bq)=(3pim)/(2Bq))`
`((3pi)(1.6xx10^I-27))/((2)(1)(1.6xx10^-19))`
`=4.712xx10^-8s`
In figure a Centre of circulr path is lying on the boundary line of magnetic field. Deviation of the particle is `180^@` and time spend in magnetic field `t=T/2`
In figure b. Centre of circular path lies outside the magnetic fiedl. Deviation of the particle is less than `180^0` and time spend in magnetic field `ltT/2`
In figure c. Centre of circular path lies inside the magnetic field. Deviation of the particle is more than `180^@` and time spent in magnetic field `tgtT/2`
`r=(mv)/(Bq)` and `omega=(Bq)/m`
a. Refer figure i
`/_CFG=90^@-theta` and `/_CEG=90^@-45^@=45^@`
since `CF=CE`
`:. /_CFG=/_CEG`
or `90^@-theta=45^@` or `theta=45^@`
Further `FG=GE=rcos45^@`
`:. EF=2FG=2rcos45^0=(2mvcos45^@)/(Bq)`
`=(2(1.6xx10^-27)(10^7)(1/sqrt2))/((1)(1.6xx10^-19))=0.14m`
b. refer figure ii In this casse particle will complete `3/4` th of circle in the magnetic field.
Hence, the time spend in the magnetic field
`t=3/4` (time period of circular motion)
`=3/4((2pim)/(Bq)=(3pim)/(2Bq))`
`((3pi)(1.6xx10^I-27))/((2)(1)(1.6xx10^-19))`
`=4.712xx10^-8s`
In figure a Centre of circulr path is lying on the boundary line of magnetic field. Deviation of the particle is `180^@` and time spend in magnetic field `t=T/2`
In figure b. Centre of circular path lies outside the magnetic fiedl. Deviation of the particle is less than `180^0` and time spend in magnetic field `ltT/2`
In figure c. Centre of circular path lies inside the magnetic field. Deviation of the particle is more than `180^@` and time spent in magnetic field `tgtT/2`
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