A charge +2q moves vertically upwards with speed v, a second charge $- q$ moves horizontally to the right with the same speed v, and a third charge + q moves horizontally to the right with the same speed v. The point P is located at a perpendicular distance a away from each charge as
shown in the figure. The magnetic field at point P is

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Into the page with magnitude $\frac{μ_{0}}{4 π} \frac{4 qv}{a^{2}}$

Into the page with magnitude $\frac{μ_{0}}{4 π} \frac{2 qv}{a^{2}}$

Out of the page with magnitude $\frac{μ_{0}}{4 π} \frac{4 qv}{a^{2}}$

Out of the page with magnitude $\frac{μ_{0}}{4 π} \frac{2 qv}{a^{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} {\vec{B}}_{P} = {\vec{B}}_{q} + {\vec{B}}_{2 q} + {\vec{B}}_{- q} \\ = \frac{μ_{0}}{4 π} \frac{qv}{a^{2}} ⊙ + \frac{μ_{0}}{4 π} \frac{2 qv}{a^{2}} ⊙ + \frac{μ_{0}}{4 π} \frac{qv}{a^{2}} ⊙ = \frac{μ_{0}}{4 π} \frac{4 qv}{a^{2}} \end{array}$