JEE Main 26-Aug-2021 Shift 1 Solved Paper

The magnetic field intensity at a distance r from centre on the axis of a coil of radius a carrying current i is given by
Baxis=

μ0

4π

2πia2

(a2+r2) 3 2

...(i)
The magnetic field intensity at the centre of the coil is given by
Bcentre=

μ0

4π

2πi

a

...(ii)
According to question, the fractional change in the magnetic fieldintensity is given by

ΔB

B

=1−

Baxis

Bcentre

...(iii)
Substituting the values of Baxis and Bcentre from Eqs. (i) and (ii) respectively in Eq. (iii), we get

ΔB

B

=1−

μ0 4π 2πia2 (a2+r2) 3 2

μ0 4π 2πi a

=1−

a3

(a2+r2) 3 2

=1−

a3

a3(1+ r2 a2 ) 3 2

=1−(1+

r2

a2

)−

3

2

Given that, r < a, applying Binomial expansion,
(1−x)−n=1−x+

2x2

2!

−

3×2x3

3!

...(−1)nxn,
=(1+nx)
we get

ΔB

B

=1−(1−

3

2

r2

a2

)

ΔB

B

=1−1+

3

2

r2

a2

ΔB

B

=

3

2

r2

a2

, which is the required expression.