AP EAPCET 23-Aug-2021 Shift 2 Paper

Given,
Side of square frame, a=1m
Electric current =I
For square, Let r be the distance of centre O from all sides, r=

a

2

Since, all sides of square produces equal field in the same direction.
Using, B=4[

µ0I

4πr

(sin‌45°+sin‌45°)]
⇒B=

4µ0I

4π( a 2 )

(

1

√2

+

1

√2

)
⇒B=

2µ0I

πa

2

√2

⇒B=

2√2µ0I

πa

...(i)
For circle, Let R be the radius of circular coil. Then,
2πR=4a
⇒R=

4a

2π

=

2a

π

Using the expression for magnetic field at centre of circular coil,
B′=

µ0I

2R

⇒B′=

µ0I

2( 2a π )

=

µ0πI

4a

...(ii)
From Eqs. (i) and (ii), we get

B

B′

=

2√2µ0I πa

µ0πI 4a

=

8√2

π2