Match List-$I$ with List-$II$.
List-$I$List-$II$
$(a)$ Magnetic Induction$(i)$ ${ML}^{2} {T}^{-2} {A}^{-1}$
$(b)$ Magnetic Flux$(ii)$ ${M}^{0} {L}^{-1} {A}$
$(c)$ Magnetic Permeability$(iii)$ ${MT}^{-2} {A}^{-1}$
$(d)$ Magnetization$(iv)$ ${MLT}^{-2} {A}^{-2}$

Choose the most appropriate answer from the options given below:

  • A
    $(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)$
  • B
    $(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)$
  • C
    $(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)$
  • D
    $(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)$

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$(a)$ What happens if a bar magnet is cut into two pieces: $(i)$ transverse to its length,$(ii)$ along its length?
$(b)$ $A$ magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet,however,experiences a force of attraction in addition to a torque. Why?
$(c)$ Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid?
$(d)$ Two identical-looking iron bars $A$ and $B$ are given,one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised? If only one is magnetised,how does one ascertain which one? [Use nothing else but the bars $A$ and $B$.]

Assertion: We cannot think of a magnetic field configuration with three poles.
Reason: $A$ bar magnet does exert a torque on itself due to its own field.

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