$d i=\frac{ q \omega}{2 \pi}$
$dB =\frac{\mu_{0} di }{2 x }$
$dB =\frac{\mu_{0} \omega \sigma}{4 \pi x } \times 2 \pi x dx$
$B =\frac{\mu_{0} \omega \sigma}{2}[ R ]$
$B =\frac{\mu_{0} \omega}{2} \times \frac{ Q }{\pi R ^{2}} \times R$
$B =\frac{\mu_{0} \omega \times Q }{2 \pi R}$
$B \propto \frac{1}{ R }$

Assertion $A:$ For measuring the potential difference across a resistance of $600\,\Omega$, the voltmeter with resistance $1000\,\Omega$ will be preferred over voltmeter with resistance $4000\,\Omega$.
Reason $R:$ Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.
In the light of the above statements, choose the most appropriate answer from the options given below.
