Q.41. An insulated wire is wound so that it forms a flat coil with N = 200 turns. The radius of the

Q: What is the correct answer for JEE Main Physics Q41?

The correct answer is B. Sol. The magnetic moment of a current loop is $M = NIA$, where N is the number of turns, I is the current, and A is the area. For a flat coil with turns distributed between $r_1$ and $r_2$, we can consider the average area. However, a more precise approach is to integrate. If the turns are uniformly distributed, the number of turns per unit radial length is $\frac{N}{r_2 - r_1}$. Consider a differ

Q: Q.41. An insulated wire is wound so that it forms a flat coil with N = 200 turns. The radius of the innermost turn is , and of the outermost turn . If 20 mA current flow in it then the magnetic…

The correct answer is B. Sol. The magnetic moment of a current loop is $M = NIA$, where N is the number of turns, I is the current, and A is the area. For a flat coil with turns distributed between $r_1$ and $r_2$, we can consider the average area. However, a more precise approach is to integrate. If the turns are uniformly distributed, the number of turns per unit radial length is $\frac{N}{r_2 - r_1}$. Consider a differ

Question

Q.41. An insulated wire is wound so that it forms a flat coil with N = 200 turns. The radius of the innermost turn is , and of the outermost turn . If 20 mA current flow in it then the magnetic moment will be . The value of a is

FirstInTest · Accepted Answer

Sol. The magnetic moment of a current loop is $M = NIA$, where N is the number of turns, I is the current, and A is the area. For a flat coil with turns distributed between $r_1$ and $r_2$, we can consider the average area. However, a more precise approach is to integrate. If the turns are uniformly distributed, the number of turns per unit radial length is $\frac{N}{r_2 - r_1}$. Consider a differential ring of thickness dr at radius r. The number of turns in this ring is $dN = \frac{N}{r_2 - r_1} dr$. The area of this ring is $dA = \pi r^2$. The magnetic moment of this differential ring is $dM = I dN dA = I \frac{N}{r_2 - r_1} dr \pi r^2$. This is incorrect. The magnetic moment of a single turn is $I \pi r^2$. So, for N turns, if they were all at the same radius, it would be $N I \pi r^2$. For a flat coil with turns distributed from $r_1$ to $r_2$, the average area is not simply $\pi (r_1+r_2)^2/4$. The magnetic moment is $M = I \sum A_i = I \sum \pi r_i^2$. If the turns are uniformly distributed, the average radius is $(r_1+r_2)/2$. The average area would be $\pi (r_1+r_2)^2/4$. So, $M = N I \pi \left(\frac{r_1+r_2}{2}ight)^2$. This is an approximation. A more accurate method for a flat spiral coil is $M = \frac{N I \pi (r_1^2 + r_1 r_2 + r_2^2)}{3}$. Let's use the average radius approach as it's simpler and often used for such problems. Average radius $r_{avg} = \frac{r_1+r_2}{2} = \frac{3+6}{2} = 4.5 	ext{ cm} = 0.045 	ext{ m}$. Area $A = \pi r_{avg}^2 = \pi (0.045)^2 	ext{ m}^2$. $N = 200$, $I = 20 	ext{ mA} = 20 	imes 10^{-3} 	ext{ A}$. $M = N I A = 200 	imes (20 	imes 10^{-3}) 	imes \pi (0.045)^2 = 4 	imes \pi 	imes (0.045)^2 = 4 	imes 3.14159 	imes 0.002025 \approx 0.025447 	ext{ A.m}^2$. We need $a 	imes 10^{-2} 	ext{ A.m}^2$. So $a = 2.5447$. This is close to 2.64. Let's try the formula $M = \frac{N I \pi (r_1^2 + r_1 r_2 + r_2^2)}{3}$. $r_1 = 0.03 	ext{ m}$, $r_2 = 0.06 	ext{ m}$. $M = \frac{200 	imes (20 	imes 10^{-3}) 	imes \pi ((0.03)^2 + (0.03)(0.06) + (0.06)^2)}{3} = \frac{4 	imes \pi 	imes (0.0009 + 0.0018 + 0.0036)}{3} = \frac{4 	imes \pi 	imes 0.0063}{3} = 4 	imes \pi 	imes 0.0021 = 0.0084 \pi \approx 0.026389 	ext{ A.m}^2$. So $a \approx 2.64$. This matches option (2).

Answer

4

Answer

64

Answer

25

Answer

2

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