Question

At an instant electron is moving in the xy plane the components of its velocity along x and y axes are 5 into 10 power 5 metre per second in and 3 into 10 power 5 metre per second respectively a magnetic field of 0.8 tesla is in the positive positive x axis at this instant the magnitude of the magnetic force on the electron is

Answer 1

Answer:

Force on the electron moving in magnetic field is given as

$F = 3.84 \times 10^{-14} \hat k$

Explanation:

velocity of the electron is given as

$\vec v = 5\times 10^5 \hat i + 3\times 10^5 \hat j$

now we have

$\vec B = 0.8 T\hat i$

now we know that magnetic force on moving charge is given as

$F = q(\vec v \times \vec B)$

so we have

$F = -(1.6 \times 10^{-19})(5\times 10^5 \hat i + 3\times 10^5 \hat j)\times (0.8\hat i)$

$F = 3.84 \times 10^{-14} \hat k$

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Topic : Magnetic force on moving charge

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