Question

Two circular coils 1 and 2 are made from the same wire but the radius of the $1^{st}$ coil is twice that of the $2^{nd}$coil. What potential difference in volts should be applied across them so that the magnetic field at their centres is the same-

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

4

Answer 2

Let $r _{1}$ and $r_{2}$ are the radius of coil 1 & 2.
If $B _{1}$ and $B _{2}$ are magnetic induction at their centre, then
$B _{1}=\frac{\mu_{0} I _{1}}{2 r _{1}}$ ; and $B _{2}=\frac{\mu_{0} I _{2}}{2 r _{2}}$
Since $B _{1}= B _{2}$ ; and $r _{1}=2 r _{2}$
therefore $I _{1}=2 I _{2}$
Again if $R _{1}$ and $R _{2}$ are resistance of the coil 1 and 2 then
$$R _{1}=2 R _{2}( as\ R \propto length =2 \pi r )$and if$V _{1}$and$V _{2}$are the potential difference across them respectively, then$\frac{ V _{1}}{ V _{2}}=\frac{ I _{1} R _{1}}{ I _{2} R _{2}}$$=\frac{\left(2 I _{2}\right)\left(2 R _{2}\right)}{ I _{2} R _{2}}=4$
Therefore, the correct option is (C) : 4.