TG EAMCET 2 May 2025 Shift 2 Paper

The length of the wire L is constant for both coils
For coil A,L=NA×2πRA
For coil B,L=NB×2πRB
‌NA×2πRA=NB×2πRB
‌NARA=NBRB
So, ‌‌‌

RA

RB

=‌

NB

NA

Magnetic field at the centre of coil‌A
BA=‌

µ0NAI

2RA

Magnetic field at the centre of coil B
BB=‌

µ0NBI

2RB

The ratio is ‌

BA

BB

=‌

‌ µ0NAI 2RA

µ0NBI 2RB

‌

BA

BB

‌=‌

NA

NB

×‌

RB

RA

=(‌

NA

NB

)⋅(‌

NA

NB

)=(‌

NA

NB

)2
‌

BA

BB

‌=(‌

2

3

)2=‌

4

9

The ratio of the magnetic fields at the centres of coils A and B is 4:9.