(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 $J$)

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$0.2$
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$2.6$
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$0.4$
D
$6.2$

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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 $J$)

Similar Questions

$A$ small bar magnet of magnetic moment $M$ is placed in a uniform magnetic field $H$. If the magnet makes an angle of $30^{\circ}$ with the field,the torque acting on the magnet is:

$A$ bar magnet of magnetic moment $1.5 \, J/T$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment perpendicular to the field direction? (in $J$)

Torques $\tau_1$ and $\tau_2$ are required for a magnetic needle to remain perpendicular to the magnetic fields of $B_1$ and $B_2$ at two different places. The ratio of $B_1: B_2$ is equal to

The magnitude of the axial field due to a bar magnet at a distance of $1 \ m$ is found to be $5 \times 10^{-8} \ T$. The magnetic moment of the bar magnet is $\left(\mu_0 = 4 \pi \times 10^{-7} \ T \ m/A\right)$. (in $A \ m^2$)

$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

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