Question

The angle of dip at a place is 60^o. A magnetic needle oscillates in a horizontal plane at this place with period T. The same needle will oscillate in a vertical plane coinciding with the magnetic meridian with a period

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

Answer 1

Answer: (D)

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

Answer 2

When needle oscillates in horizontal plane

Then it's time period is $T = 2 \pi \sqrt { \frac { I } { M B _ { H } } }$ ......(i)

When needle oscillates in vertical plane i.e. It oscillates in total earth's total magnetic field (2)

Hence $T ^ { \prime } = 2 \pi \sqrt { \frac { I } { M } }$ .....(ii)

Dividing equation (ii) by (i)

$\frac { T ^ { \prime } } { T } = \sqrt { \frac { B _ { H } } { B } } = \sqrt { \frac { B \cos \phi } { B } } = \sqrt { \cos 60 } = \frac { 1 } { \sqrt { 2 } }$⇒