Hint : \\bullet For an infinitely long wire, the magnetic field is given as B=\\frac {\\mu_o I}{2\\pi r} \\bullet Mutual inductance is given as M=\\frac {\\phi}{I} (a) \ extbf{Step 1 : Magnetic field due to infinite current carrying wire at distance x} The magnetic field of the infinite wire is- \\mathrm{B}=\\frac{\\mu_{0}}{2 \\pi} \\frac{\\mathrm{i}}{\\mathrm{x}} \ extbf{Step2:Magnetic flux through an element} Considering an elemental length of width dx at a diatance of x from the straight wire.Area of element dS=adxThe magnetic field at dx is, \\mathrm{B}_{\\mathrm{x}}=\\frac{\\mu_{0}}{2 \\pi} \\frac{\\mathrm{i}}{\\mathrm{x}} The magnetic flux through the region of width dx is, \\Rightarrow \\mathrm{d} \\phi=\\left(\\mathrm{B}_{\\mathrm{x}}\\right) \\mathrm{d} \\mathrm{S} \\Rightarrow d\\phi=\\left(\\frac{\\mu_{0}}{2 \\pi} \\frac{\\mathrm{i}}{\\mathrm{x}}\\right)(\\mathrm{adx}) \ extbf{Step3:Total Flux through square loop} Taking integration on RHS from a to 2a we have, \\Rightarrow \\phi=\\int_{\\mathrm{a}}^{2 \\mathrm{a}} \\mathrm{d} \\phi \\Rightarrow \\phi=\\frac{\\mu_{0} \\mathrm{ia}}{2 \\pi} \\ln 2 \ extbf{Step4:The mutual inductance between the long wire and square loop} M=\\frac{\\phi}{I} \\Rightarrow M=\\frac{\\mu_{0} \\mathrm{a}}{2 \\pi} \\ln 2 (c) Wire produces ⨀ magnetic field over the loop. if the loop is brought closer to the wire ⨀ magnetic field passing through the loop increases. Hence induced current produces ⨂ magnetic field so, induced current is clockwise.Therefore the correct options are both 'A' & 'C' is ( A ) The mutual inductance between the two is \\frac{\\mu_{0} a}{2 \\pi} \\ln (2) . ( C ) If a constant current is passed in the straight wire in upward direction and loop is brought close to the wire then induced current in the loop is clockwise