Question

A steady current i flows in a small square loop of wire of side L in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let $\overrightarrow { \mu _ { 1 } }$ and $\overrightarrow { \mu _ { 2 } }$ respectively denote the magnetic moments due to the current loop before and after folding. Then

• A. $\overrightarrow { \mu _ { 2 } } = 0$
• B. $\overrightarrow { \mu _ { 1 } }$ and$\overrightarrow { \mu _ { 2 } }$are in the same direction
• C. $\frac { \left| \overrightarrow { \mu _ { 1 } } \right| } { \left| \overrightarrow { \mu _ { 2 } } \right| } = \sqrt { 2 }$
• D. $\frac { \left| \overrightarrow { \mu _ { 1 } } \right| } { \left| \overrightarrow { \mu _ { 2 } } \right| } = \left( \frac { 1 } { \sqrt { 2 } } \right)$

Answer 1

Answer: (C)

$\frac { \left| \overrightarrow { \mu _ { 1 } } \right| } { \left| \overrightarrow { \mu _ { 2 } } \right| } = \sqrt { 2 }$

Answer 2

$\longleftarrow L$ $\longleftarrow \mu _ { 1 } = i L ^ { 2 }$

M = magnetic moment due to each part

$= i \left( \frac { L } { 2 } \right) \times L = \frac { i L ^ { 2 } } { 2 } = \frac { \mu _ { 1 } } { 2 }$

∴ $\mu _ { 2 } = M \sqrt { 2 } = \frac { \mu _ { 1 } } { 2 } \times \sqrt { 2 } = \frac { \mu _ { 1 } } { \sqrt { 2 } }$Initially Finally