A uniform conducting wire of length is 24a, and resistance R is wound up as a current carrying coil in the shape of an equilateral triangle of side 'a' and then in the form of a square of side 'a'. The coil is connected to a voltage source V0. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is 1 : \sqrt y where y is ................. .

Magnetics PYQ — JEE Main 2021 · 27 Aug · Shift 1 · Q61 with Solution

View written solutionFree

Correct answer: 3

Given data

Total length of wire $=24a$
Total resistance of wire $=R$
Same wire is wound into:
- an equilateral triangular coil of side $a$
- a square coil of side $a$
Both are connected to the same voltage source $V_0$

We need the ratio of magnetic moments: $m_{\triangle} : m_{\square} = 1 : \sqrt{y}$

Formula for magnetic moment

For a coil, $m = NIA$ where:

$N$ = number of turns
$I$ = current through the coil
$A$ = area of one turn

Since the same total wire is used each time, the total resistance remains $R$ in both cases. Hence current in both cases is $I = \frac{V_0}{R}$ So current is same for both coils.

Thus, $m \propto NA$

Equilateral triangle case

Each turn has perimeter $3a$ So number of turns is $N_{\triangle} = \frac{24a}{3a} = 8$

Area of one equilateral triangle of side $a$ is $A_{\triangle} = \frac{\sqrt{3}}{4}a^2$

Therefore magnetic moment,

= 8 \cdot I \cdot \frac{\sqrt{3}}{4}a^2 = 2\sqrt{3}\,Ia^2$$ --- 4. **Square case** Each turn has perimeter $$4a$$ So number of turns is $$N_{\square} = \frac{24a}{4a} = 6$$ Area of one square of side $a$ is $$A_{\square} = a^2$$ Therefore magnetic moment, $$m_{\square} = N_{\square} I A_{\square} = 6Ia^2$$ --- 5. **Take ratio** $$m_{\triangle} : m_{\square} = 2\sqrt{3}Ia^2 : 6Ia^2$$ Cancel common terms: $$= \sqrt{3} : 3$$ Now write it in the form $1 : \sqrt{y}$. Divide both terms by $\sqrt{3}$: $$1 : \frac{3}{\sqrt{3}} = 1 : \sqrt{3}$$ So, $$\sqrt{y} = \sqrt{3} \Rightarrow y=3$$ --- 6. **Final answer** $$\boxed{3}$$

More from Magnetics

View written solutionFree

Correct answer: 3

Given data

Total length of wire $=24a$
Total resistance of wire $=R$
Same wire is wound into:
- an equilateral triangular coil of side $a$
- a square coil of side $a$
Both are connected to the same voltage source $V_0$

We need the ratio of magnetic moments: $m_{\triangle} : m_{\square} = 1 : \sqrt{y}$

Formula for magnetic moment

For a coil, $m = NIA$ where:

$N$ = number of turns
$I$ = current through the coil
$A$ = area of one turn

Since the same total wire is used each time, the total resistance remains $R$ in both cases. Hence current in both cases is $I = \frac{V_0}{R}$ So current is same for both coils.

Thus, $m \propto NA$

Equilateral triangle case

Each turn has perimeter $3a$ So number of turns is $N_{\triangle} = \frac{24a}{3a} = 8$

Area of one equilateral triangle of side $a$ is $A_{\triangle} = \frac{\sqrt{3}}{4}a^2$

Therefore magnetic moment,

= 8 \cdot I \cdot \frac{\sqrt{3}}{4}a^2 = 2\sqrt{3}\,Ia^2$$ --- 4. **Square case** Each turn has perimeter $$4a$$ So number of turns is $$N_{\square} = \frac{24a}{4a} = 6$$ Area of one square of side $a$ is $$A_{\square} = a^2$$ Therefore magnetic moment, $$m_{\square} = N_{\square} I A_{\square} = 6Ia^2$$ --- 5. **Take ratio** $$m_{\triangle} : m_{\square} = 2\sqrt{3}Ia^2 : 6Ia^2$$ Cancel common terms: $$= \sqrt{3} : 3$$ Now write it in the form $1 : \sqrt{y}$. Divide both terms by $\sqrt{3}$: $$1 : \frac{3}{\sqrt{3}} = 1 : \sqrt{3}$$ So, $$\sqrt{y} = \sqrt{3} \Rightarrow y=3$$ --- 6. **Final answer** $$\boxed{3}$$

Magnetics question

Magnetics question

More from Magnetics

Magnetics question

Magnetics question

More from Magnetics