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(Solved): At \( t=0 \) an electron crosses the positive \( y \)-axis (so \( x=0 \) ) at \( 79 \mathrm{~cm} \ ...



At \( t=0 \) an electron crosses the positive \( y \)-axis (so \( x=0 \) ) at \( 79 \mathrm{~cm} \) from the origin with velo

At \( t=0 \) an electron crosses the positive \( y \)-axis (so \( x=0 \) ) at \( 79 \mathrm{~cm} \) from the origin with velocity \( 7 \times 10^{5} \mathrm{~m} / \mathrm{s} \) in the positive \( x \)-direction. It is in a uniform magnetic field. (a) Find the maqnitude and the direction of the magnetic field that will cause the electron to cross the \( x \)-axis at \( x=79 \mathrm{~cm} \). magnitude direction (b) What work is done on the electron during this motion? \( \chi \) Units are required for this answer. (c) How lonq will the trip take from \( y \)-axis to \( x \)-axis? X Dimensionally incorrect. Please check the type or dimension of your unit.


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Given: speed of electron: = v charge mag
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