TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Given, side of cube, a=1mm=1×10−3m
Radius of circular loop, R=20cm=20×10−2m
Current in loop, I=2A
As we know that, magnetic field due to loop,
B=‌

1

2

‌

µ0I

R

where, µ0 is free space permittivity, i.e. 4π×10−7 and magnetic energy stored, E=‌

1

2

‌

B2V

µ0

where, V= volume of cube.
‌ Therefore, ‌E‌‌=‌

1

2

×‌

(µ0I∕2R)2V

µ0

‌‌=‌

1

8

×‌

4π×10−7×4

(20×10−2)2

×10−9
[∵V=a3=(10−3)3=10−9m3]
‌‌=‌

2π

400

×10−12=‌

200

400

π×10−14
‌‌=‌

3.14

2

×10−14=1.57×10−14J