(A) Given:Magnetic moment of the bar magnet,$m = 10^{-4} Am^2$Magnetic field,$B = 12 \times 10^{-3} T$Angle,$\theta = 30^{\circ}$The torque $\tau$ act

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(A) Given:Magnetic moment of the bar magnet,$m = 10^{-4} Am^2$Magnetic field,$B = 12 \times 10^{-3} T$Angle,$\theta = 30^{\circ}$The torque $\tau$ acting on a bar magnet in a uniform magnetic field is given by the formula:$\tau = mB \sin \theta$Substituting the given values:$\tau = (10^{-4} Am^2) \times (12 \times 10^{-3} T) \times \sin(30^{\circ})$Since $\sin(30^{\circ}) = 0.5$:$\tau = 10^{-4} \times 12 \times 10^{-3} \times 0.5$$\tau = 12 \times 10^{-7} \times 0.5$$\tau = 6 \times 10^{-7} Nm$

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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If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field,then the torque acting on the magnet is:

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A
$6 \times 10^{-7} Nm$
B
$6 \times 10^{-5} Nm$
C
$12 \times 10^{-7} Nm$
D
$12 \times 10^{-5} Nm$

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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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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 $A \ m^2$)

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$A$ bar magnet has a magnetic moment of $200 \text{ A m}^2$. The magnet is suspended in a magnetic field of $0.30 \text{ N A}^{-1} \text{ m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$ will be:

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If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field,then the torque acting on the magnet is:

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$A$ bar magnet of magnetic moment $2 \text{ A m}^2$ lies aligned with the direction of a uniform magnetic field of $0.3 \text{ T}$. The amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is (in $\text{ J}$)

What is the magnitude of the equatorial magnetic field due to a bar magnet of length $5.0 \ cm$ at a distance $75 \ cm$ from its midpoint? The magnetic moment of the bar magnet is $0.75 \ A \ m^2$.

The dipole moment of a short bar magnet is $1.25 \, A-m^2$. The magnetic field on its axis at a distance of $0.5 \, m$ from the centre of the magnet is:

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 $A \ m^2$)

$A$ bar magnet has a magnetic moment of $200 \text{ A m}^2$. The magnet is suspended in a magnetic field of $0.30 \text{ N A}^{-1} \text{ m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$ will be:

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