NEET-XII-Physics

36: Permanent Magnets

with Solutions - page 3
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  • #3
    A uniform magnetic field of 0.20 × 10-3 T exists in the space. Find the change in the magnetic scalar potential as one moves through 50 cm along the field.
    Ans : Given:
    Magnetic field in the space, B = 0.20 × 10-3 T
    Distance moved, `` ∆r`` = 50 cm
    We know,
    `` B=-\frac{dV}{dl}``
    `` \Rightarrow dV\mathit{=}{\mathit{\int }}_{{r}_{1}}^{{r}_{2}}\mathit{-}\stackrel{\mathit{\rightharpoonup }}{B\mathit{.}}\,\mathrm{\,d\,}l``
    `` ``
    Since the magnetic field is uniform, it can come out of the integration sign.
    `` \Rightarrow ∆V\mathit{=}\mathit{-}B\mathit{.}(∆r)``
    `` ``
    Here, `` ∆V`` is the change in the potential.
    ∴ Change in the potential = -0.2 × 10-3 × 0.5
    = -0.1 × 10-3 T-m
    Here, the negative sign shows that the potential decreases.
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