Question

When a proton is released from the rest in a room, it starts with an initial acceleration $a_0$ towards north with a speed $v_0$ it moves with an initial acceleration $3a_0$ towards west. The electric and magnetic fields in the room are

• A. $\frac{m{a}_0}{e}$ west, $\frac{2m{a}_0}{e{v}_0}$ up
• B. $\frac{m{a}_0}{e}$ west, $\frac{2m{a}_0}{e{v}_0}$ down
• C. $\frac{m{a}_0}{e}$ west, $\frac{3m{a}_0}{e{v}_0}$ up
• D. $\frac{m{a}_0}{e}$ east, $\frac{3m{a}_0}{e{v}_0}$ down

Answer 1

The correct option is B $\frac{m{a}_0}{e}$ west, $\frac{2m{a}_0}{e{v}_0}$ down

$a_0=\frac{eE}{m}\Rightarrow E=\frac{m{a}_0}{e}$

$\frac{e{v}_{0}B+eE}{m}=3a_0$

$e{v}_{0}B=3m{a}_0-eE$

$=3m{a}_0-m{a}_0$

$=2m{a}_0$

$B=\frac{2m{a}_0}{e{v}_0}$