(C) The magnetic field strength $B$ at a point on the equatorial line of a bar magnet is given by the formula:$B = \frac{\mu_0}{4\pi} \frac{M}{(r^2 +

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(C) The magnetic field strength $B$ at a point on the equatorial line of a bar magnet is given by the formula:$B = \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}}$Given values:Magnetic moment $M = 4\,J\,T^{-1}$Distance from center $r = 200\,cm = 2\,m$Length of magnet $2l = 6\,cm$,so $l = 3\,cm = 0.03\,m$Since $r \gg l$,we can approximate the formula as:$B \approx \frac{\mu_0}{4\pi} \frac{M}{r^3}$Substituting the values:$B = 10^{-7} \times \frac{4}{2^3}$$B = 10^{-7} \times \frac{4}{8}$$B = 0.5 \times 10^{-7}\,T = 5 \times 10^{-8}\,T$Thus,the magnetic field strength is $5 \times 10^{-8}\,T$.

Class 12 Physics Magnetism and Matter Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.

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A
$4\times 10^{-8}\,T$
B
$3.5\times 10^{-8}\,T$
C
$5\times 10^{-8}\,T$
D
$3\times 10^{-8}\,T$

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Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.

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$A$ bar magnet with a magnetic moment $5.0\,Am^2$ is placed in a parallel position relative to a magnetic field of $0.4\,T$. The amount of work done in turning the magnet from a parallel to an antiparallel position relative to the field direction is $.........\,J$.

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