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use (i) the Ampere's law for H and (ii) ...

use (i) the Ampere's law for H and (ii) continuity of lines of B, to conclued that inside a bar magnet, (a) lines of H run from the N pole to S pole, while (b) lines of B must run from the S pole to N pole.

Text Solution

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Consider a magnetic field line of `vecB` through the bar magnet as shown in figure. It must be a closed loop. Let C be the amperian loop. Then
`int_(Q)^(P) vecH.dvecl=int_(Q)^(P)(vecB)/(mu_0).dvecl gt 0` (i.e., positive)
It is so because the angle between `vecB` and `dvecl` is less than `90^@` inside the bar magnet. So `vecB.dvecl` is positive. Hence, the lines of `vecB` must run from S pole to N pole inside the bar magnet.
According to Ampere's law, `oint_(PQR) vecH.dvecl=0 :. oint_(PQR)vecH.dvecl=int_(P)^(Q) vecH. dvecl=int_(Q)^(P) vecH.dvecl=0`
As `int_(Q)^(p) vecH.dvecl gt 0`, so `int_(P)^(Q)vecH.dvecl lt 0` (i.e., negative)
It will be so if angle between `vecH` and `dvecl` is more than `90^@`, so that `cos theta` is negative. It means the line of `vecH` must run from N pole to S pole inside the bar magnet.
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Use i the Ampere's law for H and ii continuity of lines of B, to conclude that inside a bar magnet, (a) lines of H run from the N-pole to S-pole while (b) lies of B must run from the S-pole to N -pole.

Use (i) the Ampere's law for vecH and (ii) continuity of lines of vecB , to conclude that inside a bar magnet, (a) lines of vecH run from the N pole to S pole, while (b) lines of vecB must run from the S pole to N pole

Knowledge Check

  • When the N-pole of a bar magnet points towards the south and S-pole towards the north, the null points are at the

    A
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    C
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    D
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  • When the N -pole of a bar magnet points towards the south and S-pole towards the north, the null points are at the

    A
    Magnetic axis
    B
    Magnetic centre
    C
    Perpendicular divider of magnetic axis
    D
    N and S poles
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