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

The Correct Option is C

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.