Match the entries of column I with the entries of column II in the following table.

Column I Column II

$(A)$ $(P)$ Wires $1$ and $2$ attract each other.

$(B)$ $(Q)$ Wires $1$ and $2$ do not exert any force on each other.

$(C)$ $(R)$ Magnetic fields at point $P$ due to wires $1$ and $2$ are perpendicular to each other.

$(D)$
Wires $1$ and $2$ are in the same plane without touching each other. $(S)$ Magnetic fields at point $P$ due to wires $1$ and $2$ are in the same direction.

Which of the following option has the correct combination, considering column I and column II?

A \to P, Q

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B \to Q, R

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C \to R, S

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D \to P, R

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Solution

The correct option is B $B \to Q, R$
In case $(A)$

Due to $I_{1}$ , $\to B$ at $P$ , from right-hand thumb rule, is upward and due to $I_{2}$ is out of the plane.

So, magnetic fields at point $P$ due to wires $1$ and $2$ are perpendicular to each other.

Also, $I_{1} ⊥ I_{2}$ . So, wires $1$ and $2$ do not exert any force on each other.

$∴ A \to R, Q$

In case $(B)$
Similarly, as above, $\to B$ at $P$ , due to $I_{1}$
is toward left and due to $I_{2}$ is into the plane.

So, magnetic fields at point $P$ due to wires $1$ and $2$ are perpendicular to each other.

Also, $I_{1} ⊥ I_{2}$ . So, wires $1$ and $2$ do not exert any force on each other.

$∴ B \to R, Q$

In case $(C)$

$\to B$ at $P$ , due to $I_{1}$
is toward right and due to $I_{2}$ is out of the plane.

So, magnetic fields at point $P$ due to wires $1$ and $2$ are perpendicular to each other.

Also, $I_{1} ⊥ I_{2}$ . So, wires $1$ and $2$ do not exert any force on each other.

$∴ C \to R, Q$

In case $(D)$

$\to B$ at $P$ , due to $I_{1}$ and $I_{2}$ is out of the plane.

So, magnetic fields at point $P$ due to wires $1$ and $2$ are in the same direction.

Also, $I_{1} ⊥ I_{2}$ . So, wires $1$ and $2$ do not exert any force on each other.

$∴ D \to S, Q$

Hence, option $(B)$ is correct.

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Similar questions

Column I	Column II
$(A)$ Magnetic field due to infinite long wire	$(P)$ $B = \frac{μ_{0} I}{2 π d}$
$(B)$ Magnetic field due to current element	$(Q)$ $B = \frac{μ_{0} I N}{2 R}$
$(C)$ Magnetic field at centre of current carrying coil	$(R)$ $B = \frac{μ_{0} I N R^{2}}{2 (x^{2} + R^{2})^{3 / 2}}$
$(D)$ Magnetic field at axis of coil	$(S)$ $d B = \frac{μ_{0}}{4 π} \frac{I d l sin θ}{r^{2}}$

P

(0, 0, d)

List -I	List -I
(i) Tetragonal and hexagonal	(P) $a = b \neq c$
(ii) Cubic and Rhombohedral	(Q) $α = β = γ$
(iii) Monoclinic and triclinic	(R) $a \neq b \neq c$
(iv) Orthorhombic	(S) $a = b = c$

List -I	List -I
(i) Tetragonal and hexagonal	(P) $a = b \neq c$
(ii) Cubic and Rhombohedral	(Q) $α = β = γ$
(iii) Monoclinic and triclinic	(R) $a \neq b \neq c$
(iv) Orthorhombic	(S) $a = b = c$

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Column I	Column II
$(A)$	$(P)$ Wires $1$ and $2$ attract each other.
$(B)$	$(Q)$ Wires $1$ and $2$ do not exert any force on each other.
$(C)$	$(R)$ Magnetic fields at point $P$ due to wires $1$ and $2$ are perpendicular to each other.
$(D)$ Wires $1$ and $2$ are in the same plane without touching each other.	$(S)$ Magnetic fields at point $P$ due to wires $1$ and $2$ are in the same direction.