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In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^(-4) T) is maintained. An electron is shot into the field with a speed of 4.8 x× 10^6 m s^(-1) normal to the field. Explain why thepath of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 x× 10^(-19) C, m_e = 9.1x×10^(-31) kg) |
Answer» Solution :1.  Consider a uniform magnetic field `vecB`, WHOSE field lines are perpendicularly outward from the plane of figure shown by symbol "." in above figure. 2. Consider an electron entering above magnetic field perpendicularly with velocity `vecv_(1)` at point `A_(1)`. At this point magnetic force acting on this electron is `vecF_(1)=e(vecBxxvecv_(1))` which is in the direction `vec(A_(1)C)`. This force deflects this electron along curved path from `A_(1)` to `A_(2)`. 3. At point `A_(2)`, magnetic force exerted on electron is `vecF_(3)=e(vecBxxvecv_(3))` which is in the direction `vec(A_(3)C)`. Now, if same magnetic field is extended on the left side of above magnetic field then as explained earlier, electron will complete a circle of radius r and then it will repeat this circular motion with constant speed. 4. Here, from each point on this circular path, direction of magnetic force exerted on electron is FOUND to be centripetal because it is radially inward pointing towards point C which is centre of circle. Thus, under the influence of this centripetal force, electron follows circular trajectory. 5. Expression of radius of circular path : When a particle with charge q performs circular motion in a plane PERPENDICULAR to uniform magnetic field `vecB`, necessary centripetal force is provided in the form of magnetic force acting on it. Hence, `F_(m)=F_(c)` `thereforeBqv_(_|_)=(mv_(_|_)^(2))/r`  (Where `v_(_|_)=vsintheta` is the component of velocity of particle perpendicular to magnetic field, v = constant speed of particle, `theta` = angle between initial velocity `vecv` and magnetic field `vecB`). `thereforer=(mv_(_|_))/(Bq)` `thereforer=(mvsintheta)/(Bq)` 6. At present `vecv_|_vecBand" so "theta=90^(@)`. Substituting other VALUES, `r=((9.1xx10^(-31))(4.8xx10^(6))SIN90^(@))/((6.5xx10^(-4))(1.6xx10^(-19)))` `thereforer=4.2xx10^(-2)m` |
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