(D) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the change in potential energy:$W = U_f - U_i$Potential energy $U

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(D) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the change in potential energy:$W = U_f - U_i$Potential energy $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$Initial orientation: $\vec{M}$ is along $\vec{B}$,so $\theta_i = 0^{\circ}$.$U_i = -MB \cos 0^{\circ} = -MB$Final orientation: $\vec{M}$ is perpendicular to $\vec{B}$,so $\theta_f = 90^{\circ}$.$U_f = -MB \cos 90^{\circ} = 0$Work done $W = 0 - (-MB) = MB$Given: $N = 200$,$I = 100 \mu\text{A} = 100 \times 10^{-6} \text{ A}$,$A = 2.5 \times 10^{-4} \text{ m}^2$,$B = 1 \text{ T}$.Magnetic moment $M = NIA = 200 \times (100 \times 10^{-6}) \times (2.5 \times 10^{-4}) = 5 \times 10^{-6} \text{ A m}^2$.Work $W = MB = (5 \times 10^{-6}) \times 1 = 5 \times 10^{-6} \text{ J} = 5 \mu\text{J}$.

Class 12 Physics Moving Charges and Magnetism Torque , Potential Energy and Work Done in Mangetic Field

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$A$ circular coil having $200$ turns,$2.5 \times 10^{-4} \text{ m}^2$ area and carrying $100 \mu\text{A}$ current is placed in a uniform magnetic field of $1 \text{ T}$. Initially,the magnetic dipole moment $(\vec{M})$ was directed along $\vec{B}$. The amount of work required to rotate the coil through $90^{\circ}$ from its initial orientation such that $\vec{M}$ becomes perpendicular to $\vec{B}$ is . . . . $\mu\text{J}$.

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$8$
B
$3$
C
$4$
D
$5$

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Moving Charges and Magnetism

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Torque , Potential Energy and Work Done in Mangetic Field

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$A$ single circular loop of radius $1.00 \, m$ carries a current of $10.0 \, mA$. It is placed in a uniform magnetic field of magnitude $0.500 \, T$ that is directed parallel to the plane of the loop as shown in the figure. The magnitude of the torque exerted on the loop by the magnetic field is:

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$A$ small coil of $N$ turns has an effective area $A$ and carries a current $I$. It is suspended in a horizontal magnetic field $\overrightarrow{B}$ such that its plane is perpendicular to $\overrightarrow{B}$. The work done in rotating it by $180^\circ$ about the vertical axis is

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$A$ square loop of side $l$ is kept in a uniform magnetic field $B$ such that its plane makes an angle $\alpha$ with $\vec{B}$. The loop carries a current $i$. The torque experienced by the loop in this position is

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$A$ coil in the shape of an equilateral triangle of side $2 \,cm$ is suspended from a vertex such that it hangs in a vertical plane between the poles of a permanent magnet producing a horizontal magnetic field of $100 \times 10^{-3} \,T$. The magnetic field is parallel to the plane of the coil. For the moment of couple acting on the coil to be $2 \sqrt{3} \times 10^{-5} \,Nm$, the current to be passed through the coil is (in $A$)

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$A$ current carrying circular loop of radius $2 \text{ cm}$ with unit normal $\hat{n} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$ is placed in a magnetic field $\vec{B} = B_0(3\hat{i} + 2\hat{k})$. If $B_0 = 4 \times 10^{-3} \text{ T}$ and current $I = 100\sqrt{2} \text{ A}$,the torque experienced by the loop is . . . . . . $\text{N}\cdot\text{m}$. $(\pi = 3.14)$

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$A$ single circular loop of radius $1.00 \, m$ carries a current of $10.0 \, mA$. It is placed in a uniform magnetic field of magnitude $0.500 \, T$ that is directed parallel to the plane of the loop as shown in the figure. The magnitude of the torque exerted on the loop by the magnetic field is:

$A$ small coil of $N$ turns has an effective area $A$ and carries a current $I$. It is suspended in a horizontal magnetic field $\overrightarrow{B}$ such that its plane is perpendicular to $\overrightarrow{B}$. The work done in rotating it by $180^\circ$ about the vertical axis is

$A$ square loop of side $l$ is kept in a uniform magnetic field $B$ such that its plane makes an angle $\alpha$ with $\vec{B}$. The loop carries a current $i$. The torque experienced by the loop in this position is

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