QJEEAdv07Ph5ffc: JEE Advanced 2007 Physics - Class 12 JEEAdv07Ph5ffc Solution: Two wires each carrying a steady current I are shown in four…

We can find the direction of magnetic field at $$P$$ due to a wire by using thumb rule, and the direction of force on a wire due to other is calculated by the direction of cross product of direction of current and direction of magnetic field, $$\overrightarrow{f}=q(\overrightarrow{l}\times \overrightarrow{B})$$

(A) From thumb rule, the magnetic field at point $$\mathrm{P}$$ due to both wires is in opposite direction. Since $$\mathrm{P}$$ is at midpoint, net magnetic field at $$P$$ is zero. The direction of force on one wire is towards another (from above equation), hence they attract each other. Thus, option $$(\mathrm{Q})$$ are $$(\mathrm{R})$$ are correct.

(B) Direction of magnetic field at $$\mathrm{P}$$ is same due to both the loop. Hence option $$(\mathrm{P})$$ is correct. If both the loop is very close they might act as single (like solenoid), so we cannot predict loop will attract nor repel.

(C) Similar to case $$A$$, net $$B$$ at point $$\mathrm{P}$$ is zero. And direction of force is towards another wire which means attraction. Hence, option (Q) and (R) are correct.

(D) Magnetic field at the center of current carrying loop is given by $$\mathrm{B}=\frac{{\propto }_{0}i}{2R}$$

Current in the two loop is in opposite direction so magnetic field is in opposite direction due to both the loops. Radius of two loops is different, so it will have non zero field at point $$P$$.