Circular motions of charged particles

Electricity and Magnetism Level 3

The above shows the paths of two charged particles $A$ and $B$ with the same mass after they enter a uniform magnetic field, where they travel along their respective circular orbits. Both $A$ and $B$ enter and exit the magnetic field at right angles at the same time. If the quantity of the electric charge and the kinetic energy of the charged particle $A$ are $+q$ and $E_0,$ respectively, then which of following could be the quantity of the electric charge and the kinetic energy, respectively, of the charged particle $B?$

-\frac{q}{2}

\frac{E_0}{4}

-\frac{q}{2}

\frac{E_0}{2}

-q

\frac{E_0}{2}

q

\frac{E_0}{4}

2 solutions

Ayush Gupta
Mar 27, 2014

r=p/bq;;;;,,,,,,,,,,,,,,,,p^2=2Em.....

Jayant Kumar
Mar 20, 2014

Lets define (~) as an abbreviation of ( is proportional to)

Symbols have their usual and relevant meanings.

Equating force : MVsq/R = QVB, therefore angular velocity W = (V/R) = (Q/M)B

W is a direct measure of rapidity or quickness since R is same. [W~V], and [W = (QB/M)]

since M and B are same for both, [W ~ Q]

Given that A covers twice as much of angle in same time, W(A):W(B) = (2:1),

therefore Q(A):Q(B) = W(A):W(B) = 2:1

Trajectories are opposite, thereby Q(B) ~ (Minus)Q(A) ie. Q(B) = (-q/2)

Now, [W ~ V], then [Wsq ~ Vsq]. Since mass is same, [E~Vsq], that is, [E~Wsq].

therefore E(A):E(B) = [W(A):W(B)]sq = 4:1

thereby E(B) = (E0/4)

Circular motions of charged particles

2 solutions

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