AP EAPCET 23-Aug-2021 Shift 1 Paper

Given, R=5‌cm=0.05m
r=12‌cm=012m
B=250µT =250×10−6T
The magnetic field due to a current carrying circular loop at its axial line is B=

1

2(R2+r2)3∕2

Here, n=1 ,
⇒B=

1

2

µ0IR2

(R2+r2)3∕2

...(i)
The magnetic field at the centre of loop is
B0=

1

2

µ0I

R

Dividing Eq. (ii) by Eq. (i), we get

B0

B

=

µ0I

2R

×

2R2+r2)3∕2

µ0IR2

=

(R2+r2)3∕2

R3

⇒B0=B×

(R2+r2)3∕2

R3

=250×10−6×

[(0.05)2+(012)2]3∕2

(0.05)3

=250×10−6×

(0.0169) 3 2

(0.05)3

=4394µT