Question

Determine the magnitude and direction of the magnetic force on the particle when it moves
in a uniform magnetic field of 0.8 T? The mass of the particle is 5.00x10-kg and the
charge of the particle is 3.50x10-8 C. The velocity of the particle is 2.00 x10 m/s, and is
pointed in positive y-direction. Consider the magnetic field is in the negative x-direction.
(a) 5.60x10- N, - z-direction
(b) 5.60x10-° N, +z-direction
(c) 0
(d) None of the above​

Answer 1

Answer:

Force on the moving charge is given as

$F = 5.60 \times 10^{-7}$ + z direction

Explanation:

As we know that the magnetic force on moving charge is given as

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

here we have

$q = 3.50 \times 10^{-8} C$

$v = 2 \times 10 m/s \hat j$

$B = -0.8 \hat i$

now we have

$F = (3.50 \times 10^{-8})(2 \times 10)(0.8) (\hat j \times (-\hat i))$

$F = 5.60 \times 10^{-7} \hat k$

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Topic : Magnetic force on moving charge

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