Question

In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth's magnetic field is 2 sec. When a magnet is brought near and parallel to it, the time period reduces to 1 sec. The ratio H/F of the horizontal component H and the field F due to magnet will be

• A. 3
• B. $\frac { 1 } { 3 }$
• C. $\sqrt { 3 }$
• D. $\frac { 1 } { \sqrt { 3 } }$

Answer 1

Answer: (B)

$\frac { 1 } { 3 }$

Answer 2

Time period decreases i.e. field due to magnet (F) assist the horizontal component of earth's magnetic field (see theory)

Hence by using $\frac { B } { B _ { H } } = \left( \frac { T } { T ^ { \prime } } \right) - 1$ ⇒

$\frac { F } { H } = \left( \frac { 2 } { 1 } \right) ^ { 2 } - 1 = 3 \Rightarrow \frac { H } { F } = \frac { 1 } { 3 }$.