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An electron is moving vertically upwards in the uniform magnetic field of the earth . The velocity of the electron is $2 .0 \times 10^{6} {ms}^{- 1}$ and the horizontal component of earth’s magnetic field is $0 .32 \times 10^{- 4} tesla$ . Obtain the magnetic and direction of the force on the electron. $[e = 1 .6 \times 10^{- 19} C]$

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$1 .024 \times 10^{- 17} N; East wards$

$3 .04 \times 10^{10} N; North wards$

$1 .024 \times 10^{- 17} N; South wards$

$2 .02 \times 10^{6} N; west wards$

answer is B.

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Detailed Solution

The magnetic field is directed horizontally northwards and the direction of motion . The +ve charge (current ) is vertically downwards . By Fleming’s left hand rule , the force on the electron will be directed east wards

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