Question

A proton moves through a region containing a uniform electric field given by E = 32.0 ĵ V/m and a uniform magnetic field B = (0.200 î + 0.300 ĵ + 0.400 k) T. Determine the acceleration of the proton when it has a velocity v = 230 î m/s.

Answer 1

The acceleration of the proton is (-5.748 x 10⁹ j + 6.61 x 10⁹ k) m/s²

Step-by-step explanation:

Lorentz force acting on the proton is given as;

F = ma = qE + qvB

ma = q(E + v x B)

`a = (q)/(m) (E + V \ X \ B)`

where;

a is the acceleration of the proton

q is the charge of the proton

m is the mass of the proton

E is the electric field strength

v is the velocity of the proton

B is magnetic field strength

First, find the cross product of V and B

`V \ X \ B =\left[\begin{array}{ccc}i&j&k\\230&0&0\\0.200&0.300&0.400\end{array}\right] \\\\V \ X \ B = i(0) -230(0.400) j + 230(0.300) k \\\\V \ X \ B = -92 j + 69 k`

Then, add E

E + V X B = (32.0 ĵ ) + (-92j + 69 k)

= -60 j + 69 k

Finally, substitute in this value back to the main equation and calculate 'a'

`a = (q)/(m) (E + V \ X \ B)\\\\a = (1.6 *10^(-19))/(1.67*10^(-27)) (-60 j + 69k)\\\\a = 9.58*10^7(-60 j + 69k)\\\\a = (-5.748*10^9 j + 6.61*10^9 k) \ m/s^2`

Thus, the acceleration of the proton is (-5.748 x 10⁹ j + 6.61 x 10⁹ k) m/s²