An electron is projected horizontally with kinectic energy 10 kev. A magnetic field of strength 1*10^7 T exists in the vertically upward direction. Calculate the sideways deflection of the electron in travelling through 1 metre.

Asked by rohit | 21st Jul, 2015, 02:13: AM

Expert Answer:

begin mathsize 12px style straight K. straight E. equals 10 space keV equals 1.6 cross times 10 to the power of negative 15 end exponent space straight J Magnetic space field comma space straight B equals 1 cross times 10 to the power of negative 7 end exponent space straight T Since comma space magneti space field space is space vertically space upward space hence comma space its space direction space found space by space using space right minus hand space screw space rule. Hence comma space the space electron space get space directed space towards space left. Kinetic space energy equals 1 half mv squared straight v equals square root of fraction numerator KE cross times 2 over denominator straight m end fraction end root space space space space space space space space left parenthesis straight i right parenthesis Magnetic space force space iis space given space as comma space straight F equals Bqv space space space space space left parenthesis ii right parenthesis and space straight F equals ma space space space space space left parenthesis iii right parenthesis from space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis ma equals Bqv straight a equals Bqv over straight m Applying space eqution space of space motion comma straight s equals ut plus 1 half at squared As space there space is space no space force space acting space on space the space electron space in space the space horizontal space direction comma space the space velocity space of space the space electron space remain space constant space in space this space direction. space So comma space the space time space taken space to space cross space straight a space distance space of space 1 straight m space in space the space horizontal space direction space in space the space magnetic space field space is straight t equals straight x over straight v Substituting space all space values space in space equtions space of space motion comma straight s equals 1 half cross times Bqv over straight m cross times open parentheses straight x over straight v close parentheses squared equals fraction numerator qBx squared over denominator 2 mv end fraction equals fraction numerator qBx squared over denominator 2 straight m square root of fraction numerator KE cross times 2 over denominator straight m end fraction end root space space end fraction straight s equals 1 half cross times fraction numerator 1.6 cross times 10 to the power of negative 19 end exponent cross times 1 cross times 10 to the power of negative 7 end exponent cross times 1 squared over denominator 9.1 cross times 10 to the power of negative 31 end exponent cross times square root of begin display style fraction numerator 1.6 cross times 10 to the power of negative 16 end exponent cross times 2 over denominator 9.1 cross times 10 to the power of negative 31 end exponent end fraction end style end root end fraction straight s equals 0.0148 space straight m end style

Answered by Priyanka Kumbhar | 21st Jul, 2015, 03:33: PM

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