Question

A magnetic needle is made to vibrate in uniform field H, then its time period is T. If it vibrates in the field of intensity 4H, its time period will be

• A. $2 T$
• B.
• C. $2 / T$
• D. $T$

Answer 1

Answer: (B)

Answer 2

$T = 2 \pi \sqrt { \frac { I } { M B _ { H } } } \Rightarrow \frac { T _ { 1 } } { T _ { 2 } } = \sqrt { \frac { \left( B _ { H } \right) _ { 2 } } { \left( B _ { H } \right) _ { 1 } } }$

$\Rightarrow T _ { 2 } = T \sqrt { \frac { ( B H ) _ { 1 } } { ( B H ) _ { 2 } } } = \frac { T } { 2 } \quad \left( \because \left( B _ { H } \right) _ { 2 } = 4 \left( B _ { H } \right) _ { 1 } \right)$