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CBSE-XI-Physics

35: Magnetic Field due to a Current

with Solutions -

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    Section : i
  • Qstn #1
    An electric current flows in a wire from north to south. What will be the direction of the magnetic field due to this wire at a point
    Ans : According to the right-hand thumb rule, if the thumb of our right hand points in the direction of the current flowing, then the curling of the fingers will show the direction of the magnetic field developed due to it and vice versa. Let us consider the case where an electric current flows north to south in a wire.
    According to the right-hand thumb rule,
  • #1-a
    east of the wire,
    Ans : for any point in the east of the wire, the magnetic field will come out of the plane of paper
  • #1-b
    west of the wire,
    Ans : for a point in the west of the wire, the magnetic field will enter the plane of paper
  • #1-c
    vertically above the wire and
    Ans : for any point vertically above the wire, the magnetic field will be from right to left
  • #1-d
    vertically below the wire?
    Ans : for any point vertically below the wire, the magnetic field will be from left to right
    Page No 248:
  • Qstn #2
    The magnetic field due to a long straight wire has been derived in terms of µ, i and d. Express this in terms of ε0, c, i and d.
    Ans : The magnetic field due to a long, straight wire is given by
    `` B=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}i}{2\,\mathrm{\,\pi \,}d}``
    `` \because \,\mathrm{\,Speed\,}\,\mathrm{\,of\,}\,\mathrm{\,light\,},\,\mathrm{\,c\,}=\frac{1}{\sqrt{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{\,\mathrm{\,\epsilon \,}}_{\,\mathrm{\,o\,}}}}``
    `` \Rightarrow {\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}=\frac{1}{{\,\mathrm{\,c\,}}^{2}{\,\mathrm{\,\epsilon \,}}_{\,\mathrm{\,o\,}}}``
    `` \Rightarrow B=\frac{i}{2{\,\mathrm{\,\pi c\,}}^{2}{\,\mathrm{\,\epsilon \,}}_{\,\mathrm{\,o\,}}d}``
    (In terms of ε0, c, i and d)
    Page No 248:
  • Qstn #3
    You are facing a circular wire carrying an electric current. The current is clockwise as seen by you. Is the field at the centre coming towards you or going away from you?
    Ans : According to the right-hand thumb rule, if we curl the fingers of our right hand in the direction of the current flowing, then the thumb will point in the direction of the magnetic field developed due to it and vice versa. Therefore, in this case, the field at the centre is going away from us.
    Page No 248:
  • Qstn #4
    In Ampere’s
    ∮B→·dl→=μ0i,the current outside the curve is not included on the right hand side. Does it mean that the magnetic field B calculated by using Ampere’s law, gives the contribution of only the currents crossing the area bounded by the curve?
    Ans : In Ampere's law `` \oint \stackrel{\to }{B}.\stackrel{\to }{dl}={\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}i``, i is the total current crossing the area bounded by the closed curve. The magnetic field B on the left-hand side is the resultant field due to all existing currents.
    Page No 248:
  • Qstn #5
    The magnetic field inside a tightly wound, long solenoid is B = µ0 ni. It suggests that the field does not depend on the total length of the solenoid, and hence if we add more loops at the ends of a solenoid the field should not increase. Explain qualitatively why the extra-added loops do not have a considerable effect on the field inside the solenoid.
    Ans : The magnetic field due to a long solenoid is given as B = µ0ni, where n is the number of loops per unit length. So, if we add more loops at the ends of the solenoid, there will be an increase in the number of loops and an increase in the length, due to which the ratio n will remain unvaried, thereby leading to not a considerable effect on the field inside the solenoid.
    Page No 248:
  • Qstn #6
    A long, straight wire carries a current. Is Ampere’s law valid for a loop that does not enclose the wire, or that encloses the wire but is not circular?
    Ans : Ampere's law is valid for a loop that is not circular. However, it should have some charge distribution in the area enclosed so as to have a constant electric field in the region and a non-zero magnetic field. Even if the loop defined does not have its own charge distribution but has electric influence of some other charge distribution, it can have some constant magnetic field (`` \oint \stackrel{\to }{B}.d\stackrel{\to }{l}={\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{\,\mathrm{\,enclosed\,}}``).
    Page No 248:
  • Qstn #7
    A straight wire carrying an electric current is placed along the axis of a uniformly charged ring. Will there be a magnetic force on the wire if the ring starts rotating about the wire? If yes, in which direction?
    Ans : The magnetic force on a wire carrying an electric current i is `` \stackrel{\to }{F}=i.(\stackrel{\to }{l}\times \stackrel{\to }{B})``, where l is the length of the wire and B is the magnetic field acting on it. If a uniformly charged ring starts rotating around a straight wire, then according to the right-hand thumb rule, the magnetic field due to the ring on the current carrying straight wire placed at its axis will be parallel to it. So, the cross product will be
    `` (\stackrel{\to }{l}\times \stackrel{\to }{B})=0``
    `` \Rightarrow \stackrel{\to }{F}=0``
    Therefore, no magnetic force will act on the wire.
    Page No 248:
  • Qstn #8
    Two wires carrying equal currents i each, are placed perpendicular to each other, just avoiding a contact. If one wire is held fixed and the other is free to move under magnetic forces, what kind of motion will result?
    Ans : The magnetic force on a wire carrying an electric current i is `` \stackrel{\to }{F}=i.(\stackrel{\to }{l}\times \stackrel{\to }{B})``, ​where l is the length of the wire and B is the magnetic field acting on it. Suppose we have one wire in the horizontal direction (fixed) and other wire in the vertical direction (free to move). If the horizontal wire is carrying current from right to left is held fixed perpendicular to the vertical wire, which is free to move, the upper portion of the free wire will tend to move in the left direction and the lower portion of the wire will tend to move in the right direction, according to Fleming's left hand rule, as the magnetic field acting on the wire due to the fixed wire will point into the plane of paper above the wire and come out of the paper below the horizontal wire and the current will flow in upward direction in the free wire. Thus, the free wire will tend to become parallel to the fixed wire so as to experience maximum attractive force.
    Page No 248:
  • Qstn #9
    Two proton beams going in the same direction repel each other whereas two wires carrying currents in the same direction attract each other. Explain.
    Ans : Two proton beams going in the same direction repel each other, as they are like charges and we know that like charges repel each other.
    When a charge is in motion then a magnetic field is associated with it. Two wires carrying currents in the same direction produce their fields (acting on each other) in opposite directions so the resulting magnetic force acting on them is attractive. Due to the magnetic force, these two wires attract each other.
    But when a charge is at rest then only an electric field is associated with it and no magnetic fiels is produced by it. So at rest, it repels a like charge by exerting a electric force on it.
    Charge in motion can produce both electric field and magnetic field.
    The attractive force between two current carrying wires is due to the magnetic field and repulsive force is due to the electric field.
    Page No 248:
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