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The magnetic field shown in the figure consists of two uniform regions. The width of the first part is 5cm and the magnetic induction here is `0.001` T. The width of the other part is also 5cm, with the direction of the induction being opposite in direction and `0.002` T in magnitude. What should be the minimum speed of the electron arriving from the direction indicated in the figure so that it can pass through the magnetic field? Mass of electron `=9xx10^(-31)`kg A. `8/9xx10^(7) m//s`B. `4/9xx10^(7) m//s`C. `16/9xx10^(7) m//s`D. none |
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Answer» Correct Answer - A Suppose in 1^(st) region radius of circular path is r_(1) & in region 2 this is r_(2). `:.r_(1)lgt 5 & r_(2)gt5` `r=mv//(qB)` so,`v_(min)=(rqB_(min))/m` `:.v_(min)=(5xx10^(-2)xx1.6xx10^(-19)xx0.01)/(9xx10^(-31))=8/9xx10^(7)m//s` |
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