Question

A small bar magnet A oscillates in a horizontal plane with a period T at a place where the angle of dip is 60^o. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

• A. $\frac { T } { \sqrt { 2 } }$
• B. $T$
• C. $\sqrt { 2 } T$
• D. $2 T$

Answer 1

Answer: (A)

$\frac { T } { \sqrt { 2 } }$

Answer 2

$T = 2 \pi \sqrt { \frac { I } { M B } } \Rightarrow \frac { T } { T ^ { \prime } } = \sqrt { \frac { B ^ { \prime } } { B } } = \sqrt { \frac { B } { B _ { H } } }$

$\Rightarrow \frac { T } { T ^ { \prime } } = \sqrt { \frac { 1 } { \cos \phi } } = \sqrt { \frac { 1 } { \cos 60 ^ { \circ } } } = \sqrt { 2 }$ $\Rightarrow T ^ { \prime } = \frac { T } { \sqrt { 2 } }$