Question

A magnet makes 40 oscillations per minute at a place having magnetic field intensity $B _ { H } = 0.1 \times 10 ^ { - 5 }$. At another place, it takes 2.5 sec to complete one-vibration. The value of earth's horizontal field at that place

• A.
• B.
• C. $0.66 \times 10 ^ { - 8 } T$
• D. $1.2 \times 10 ^ { - 6 } T$

Answer 1

Answer: (B)

Answer 2

By using $\frac { T _ { 1 } } { T _ { 2 } } = \sqrt { \frac { \left( B _ { H } \right) _ { 2 } } { \left( B _ { H } \right) _ { 1 } } }$⇒

$\frac { 60 / 40 } { 2,5 } = \sqrt { \frac { \left( B _ { H } \right) _ { 2 } } { 0.1 \times 10 ^ { - 5 } } }$ ⇒ $\left( B _ { H } \right) _ { 2 } = 0.36 \times 10 ^ { - 6 } T$.