Question

Work required to rotate given square frame by ${180}^{\circ }$ can be given as

• A. $\frac{{\mu }_0}{2\pi }{i}_1{i}_{2}aln\left(2\right)$
• B. $\frac{{\mu }_0}{\pi }{i}_1{i}_{2}aln\left(2\right)$
• C. $\frac{{\mu }_0}{\pi }{i}_1{i}_{2}a$
• D. $\frac{{\mu }_0}{2\pi }{i}_1{i}_{2}a$

Answer 1

The correct option is B $\frac{{\mu }_0}{\pi }{i}_1{i}_{2}aln\left(2\right)$


Work done = change in P.E
Where $P.E=\int B.d\mu =\int B.i_{2}dA=i_2\int B.dA=i_2\varphi$
So work $=i_2\varphi -(-i_2\varphi )=2i_2\varphi$
$d\varphi =\frac{{\mu }_0}{2\pi }\frac{i_1}{x}.adx$
$\varphi =\int d\varphi =\frac{{\mu }_0}{2\pi }{i}_{1}a{\int }_a^{2a}\frac{dx}{x}=\frac{{\mu }_0}{2\pi }{i}_{1}aln\left(2\right)$
$\therefore$ work $=2\frac{{\mu }_0}{2\pi }{i}_1{i}_{2}aln\left(2\right)$