# A particle having charge q = 6 μC and mass m = 0.2 grams, moves in uniform external electric and magnetic fields given by E = (2i+ 3j+ 9k)N/C and B = (3i+j+ 2k) T. If the particle is presently...

A particle having charge q = 6 μC and mass m = 0.2 grams, moves in uniform external electric and magnetic fields given by

E = (2i+ 3j+ 9k)N/C and B = (3i+j+ 2k) T.

If the particle is presently moving with velocity

v = (3i+ 2j + 4k) m/s,

determine the magnitude of the particle’s acceleration at this moment in time.

(a) 0.00 m/s2 (b) 0.11 m/s2 (c) 0.22 m/s2

(d) 0.33 m/s2 (e) 0.44 m/s2

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First, we use Lorentz Force Law:

F = qE + qV x B` `

But, then, F = m*dV/dt. So:

m*A = qE + qV x B

Here, m = mass = 0.2 grams = 2 * 10^-4 kg

V = velocity vector

A = acceleration vector

q = charge = 6*10^-6 C

E = Electric field vector

B = Magnetic field vector

So, we are looking for A. Plugging in all the numbers:

(2*10^-4)*A = (6*10^-6)(2i+ 3j+ 9k) + (6*10^-6)(3i+ 2j + 4k)(3i+j+ 2k)

Then, simplifying:

(2*10^-4)*A = (1.2*10^-5 i + 1.8*10^-5 j + 5.4*10^-5 k) + (6*10^-6)(0 i + 6 j + -3 k)

(2*10^-4)*A = (1.2*10^-5 i + 1.8*10^-5 j + 5.4*10^-5 k) + (0 i + 3.6*10^-5 j + -1.8*10^-5 k)

Then, adding the vectors on the right:

(2*10^-4)*A = (1.2*10^-5 i + 5.4*10^-5 j + 3.6*10^-5 k)

Then, dividing each side by 2*10^-4, the mass, we get the acceleration:

A = 0.6*10^-1 i + 2.7*10^-1 j + 1.8*10^-1 k

To find the magnitude, we:

- square each direction of this vector

- add those up

- take the square root of that

So:

(0.6*10^-1)^2 + (2.7*10^-1)^2 + (1.8*10^-1)^2 = 0.1089

sqrt(0.1089) = 0.33

So, the answer would be "d", 0.33 m/s2