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Physics Question 41 – JEE-MAIN 2026

An insulated wire is wound so that it forms a flat coil with N=200 turns. The radius of the innermost turn is r1=3 cm, and of the outermost turn r2=6 cm. If 20 mA current flows in it then the magnetic moment will be α×102 A.m2. The value of α is _______.

The magnetic moment of a current-carrying coil depends on the current, the number of turns, and the area enclosed by the turns.

Step 1: Identify the formula for magnetic moment of a flat spiral coil✦ Active

For a flat spiral coil with N turns uniformly distributed between an inner radius r1 and an outer radius r2, carrying a current I, the magnetic moment M is given by:

M=INπ3(r12+r1r2+r22)
Step 2: Substitute given values and calculate the magnetic moment○ Expand

Given values are: N=200 turns, r1=3 cm=0.03 m, r2=6 cm=0.06 m, and I=20 mA=20×103 A. Substitute these into the formula:

M=(20×103 A)(200)π3((0.03 m)2+(0.03 m)(0.06 m)+(0.06 m)2) M=4π3(0.0009+0.0018+0.0036) M=4π3(0.0063)=4π(0.0021)=0.0084π A.m2
💡 Teacher's Secret Hint

Ensure all units are converted to SI units before calculation to avoid errors.

Step 3: Determine the value of α○ Expand

Using the approximation π3.14159, the magnetic moment is:

M0.0084×3.141590.026389 A.m2

The problem states that the magnetic moment is α×102 A.m2. Therefore:

α×102=0.026389 α=0.026389102=2.6389

Rounding to two decimal places, α2.64. This corresponds to option 2.

💡 Teacher's Secret Hint

Pay attention to the required format of the final answer, specifically the power of 10.

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