$\cos\theta=\frac{1}{2},$
$\theta=60^\circ\ \&\ \angle\text{AOB}=60^\circ$
$\text{B}=\frac{\mu_0\text{I}}{2\pi\text{r}}=\frac{10^{-7}\times2\times10}{2\times10^{-2}}=10^{-4}\text{T}$
So net is $[(10^{-4})^2]+(10^{-4})^2+2(10^{-8})\cos60^\circ]^\frac{1}{2}$
$=10^{-4}\Big[1+1+2\times\frac{1}{2}\Big]^\frac{1}{2}$
$=10^{-4}\times\sqrt3\text{T}$
$=1.732\times10^{-4}\text{T}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
A parallel plate capacitor, each with plate area A and separation d, is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change, if any, will take place in.