Question

The radius of a circular of wire is $R$ and it carries a current of $I$ ampere. At its centre a smaller ring of radius $r$ with current $i$ and $N$ turns is placed. Assuming that the planes of two rings are perpendicular to each other and the magnetic field at the centre of bigger ring is constant then the torque acting on smaller ring will be:-

• A. $Ni\pi r^2\times \left(\frac{{\mu }_{0}I}{2R}\right)$
• B. $zero$
• C. $Ni{r}^2\times \left(\frac{{\mu }_{0}I}{2R}\right)$
• D. $Ni\pi r^2\times \left(\frac{I^2}{2R}\right)$

Answer 1

The correct option is A $Ni\pi r^2\times \left(\frac{{\mu }_{0}I}{2R}\right)$

Given: radius of a circular of wire = RR ,current = II ,smaller ring of radius= rr with current= ii ,turn=N

Solution: As we know that the formula of torque is :

$\tau =NiABsin\theta$

$\tau =Ni\pi r^2\left(\frac{{\mu }_{0}I}{2R}\right)sin{90}^0$

$\tau =Ni\pi r^2\left(\frac{{\mu }_{0}I}{2R}\right)$

Hence the correct answer is A