A particle with mass 1.30 10 -3 kg and a charge of 1.22 10 -8 C has, at a given

Question

A particle with mass 1.30 10-3 kg and a charge of 1.22 10-8 C has, at a given instant, a velocity v = (3.80 104 m/s) . What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field = (1.63 T) + (0.980 T) ?

A particle with mass 1.30 x 103 kg and a charge of 1.22 x 108 c has, at a given instant, a velocity v (3.80 x 104 m/s) j. What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field B = (1.63 T) + (0.980 T) ? ^ ^ m/s2 j+ m/s2 k

Explanation / Answer

We know that the force on the moving charge in magnetic field is given by
F = q [v x B]
So first we calculate cross product of [v x B]
[v x B] = 0 i + 0 j - 6.194*104 k
[v x B] = - 6.194*104 k
Now the force
F = 1.22*10-8*(-6.194*104) k =(- 7.557*10-4 k ) N
Now we know that the acceleration is
a = F/m = - 7.557*10-4 k / 1.3*10-3 = - 0.5813 k
Hence the acceleration magnitude is = 0.5813 m/s2
And the direction will be toward - Z axis .

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