(A) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula: $W = MB(\cos \theta_1 - \cos \theta_2)$.Here,the magn

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(A) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula: $W = MB(\cos \theta_1 - \cos \theta_2)$.Here,the magnetic moment $M = 10^4 \,J \,T^{-1}$,the magnetic field $B = 4 \times 10^{-5} \,T$,the initial angle $\theta_1 = 0^{\circ}$,and the final angle $\theta_2 = 60^{\circ}$.Substituting the values into the formula:$W = (10^4) \times (4 \times 10^{-5}) \times (\cos 0^{\circ} - \cos 60^{\circ})$$W = 0.4 \times (1 - 0.5)$$W = 0.4 \times 0.5 = 0.2 \,J$.Therefore,the work done is $0.2 \,J$.

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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$A$ short bar magnet of magnetic moment $10^4 \,J \,T^{-1}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from the direction parallel to a horizontal magnetic field of $4 \times 10^{-5} \,T$ to a direction $60^{\circ}$ to the direction of the field is (in $J$)

AP EAMCET 2025 All AP EAMCET

A
$0.2$
B
$2.6$
C
$0.4$
D
$6.2$

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Magnetism and Matter

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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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$A$ short bar magnet of magnetic moment $10^4 \,J \,T^{-1}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from the direction parallel to a horizontal magnetic field of $4 \times 10^{-5} \,T$ to a direction $60^{\circ}$ to the direction of the field is (in $J$)

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Write an expression for the magnitude of the magnetic field at a point lying on the equatorial line of a bar magnet.

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$A$ bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position is

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