MEC O15

Analysis: The magnetic induction is doubled, so the new field is $B’ = 2B$. We need to find the new displacement $S’$.

We use the magnitude relationship derived in the previous question:

$$ (mv_0)^2 = S’^2 (b^2 + q^2 B’^2) $$

Step 1: Substitute Knowns

From Q14, we know:

• $mv_0 = 10b$
• $qB = \frac{4b}{3}$

The new term $qB’ = q(2B) = 2(qB) = 2(\frac{4b}{3}) = \frac{8b}{3}$.

Step 2: Solve for S’

$$ (10b)^2 = S’^2 \left( b^2 + \left(\frac{8b}{3}\right)^2 \right) $$ $$ 100b^2 = S’^2 \left( b^2 + \frac{64b^2}{9} \right) $$

Factor out $b^2$:

$$ 100b^2 = S’^2 b^2 \left( 1 + \frac{64}{9} \right) $$ $$ 100 = S’^2 \left( \frac{9 + 64}{9} \right) $$ $$ 100 = S’^2 \left( \frac{73}{9} \right) $$ $$ S’^2 = \frac{900}{73} $$ $$ S’ = \sqrt{\frac{900}{73}} = \frac{30}{\sqrt{73}} \text{ m} $$