StemCET Logo

Physics Question 42 – JEE-MAIN 2026

← Back
A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A. The magnetic energy stored inside the cube is α×10−14 J. The value of α is _______. (μ0=4π×10−7 Tm/A, π=3.14)

The magnetic energy stored in a given volume is determined by the magnetic field strength within that volume.

Step 1: Calculate Magnetic Field at the Center of the Loop✦ Active

The magnetic field B at the center of a circular loop of radius R carrying current I is given by the formula B=μ0I2R. Given R=10 cm=0.1 m, I=2 A, and μ0=4π×10−7 Tm/A.

B=(4π×10−7 Tm/A)×(2 A)2×(0.1 m)=8π×10−70.2=40π×10−7=4π×10−6 T
Step 2: Calculate Magnetic Energy Density○ Expand

The magnetic energy density u in a region with magnetic field B is given by u=B22μ0.

u=(4π×10−6 T)22×(4π×10−7 Tm/A)=16π2×10−128π×10−7=2π×10−5 J/m3
Step 3: Calculate Total Magnetic Energy and Determine α○ Expand

The volume of the small cube is V=L3. Given side L=1 mm=1×10−3 m. So, V=(1×10−3 m)3=1×10−9 m3. The total magnetic energy U stored inside the cube is the product of energy density and volume, U=u×V.

U=(2π×10−5 J/m3)×(1×10−9 m3)=2π×10−14 J

The problem states that the magnetic energy stored is α×10−14 J. Comparing this with our calculated value, we get α=2π. Using the given value π=3.14:

α=2×3.14=6.28
💡 Teacher's Secret Hint

Ensure to use the correct units and convert them to SI before performing calculations.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.