MHTCET 13 May 2023 Shift 2 Solved Paper

The magnetic field at the centre of a current carrying loop is
B=‌

µ0

4π

(‌

2πI

a

)=‌

µ0l

2a

....(i)
and magnetic moment M is
M=I(πa2)....(ii)
Thus, ‌‌X=‌

B

M

=‌

µ0I

2a

×‌

I

Iπa2

=‌

µ0

2πa3

....(iii)
Now, according to question, when both the current and radius are doubled, then the ratio will be
⇒‌

µ0

2π(2a)3

=‌

µ0

8(2πa3)

From Eq. (iii), ‌

µ0

2πa3

=X
So, ‌

µ0

8(2πa3)

=‌

X

8