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Aseem Mahajan 4 years, 6 months ago
Confirm carefully about that khushi kamboj I highly doubt its him
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Aseem Mahajan 4 years, 6 months ago
Irms = $$\sf \dfrac{V_{rms}}{X_L}$$
Irms = $$\sf \dfrac{220}{2\pi 50 \times 44 × 10^{-3}}$$
Irms = 1.65 A
Aseem Mahajan 4 years, 6 months ago
Irms = $$\sf \dfrac{V_{rms}}{X_L}$$
Irms = $$\sf \dfrac{220}{2\pi 50 \times 44 × 10^{-3}}$$
Irms = 1.65 A
Aseem Mahajan 4 years, 6 months ago
Irms = $$\sf \dfrac{V_{rms}}{X_L}$$
Irms = $$\sf \dfrac{220}{2\pi 50 \times 44 × 10^{-3}}$$
Irms = 1.65 A
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Sia ? 4 years, 5 months ago
Consider a bar magnet NS of length 2l placed in a uniform magnetic field {tex}\vec B{/tex}. Let qm be the pole strength of its every pole. Let the magnetic axis of the bar magnet make an angle {tex}\theta{/tex} with the field {tex}\vec B{/tex}, as shown in the figure.
Force on N-pole = qmB ; along {tex}\vec B{/tex}
Force on S-pole = qm B, opposite to {tex}\vec B{/tex}
The forces on the two poles are equal and opposite. They form a couple. Moment of couple or torque is given by
{tex}\tau{/tex} = Force {tex}\times{/tex} perpendicular distance
= {tex}q_{m} B \times 2 l \sin \theta{/tex} = {tex}\left(q_{m} \times 2 l\right) B \sin \theta{/tex}
or {tex}\tau{/tex} = mB sin {tex}\theta{/tex} .....(i)
where m = qm {tex}\times{/tex} 2l, is the magnetic dipole moment of the bar magnet. In vector notation,
{tex}\vec{\tau}=\vec{m} \times \vec{B}{/tex}
As shown in figure,
When a magnetic dipole is placed in a uniform magnetic field {tex}\vec B{/tex} at angle {tex}\theta{/tex} with it, it experiences a torque
{tex}\tau{/tex} = mB sin {tex}\theta{/tex}
This torque tends to align the dipole in the direction of {tex}\vec B{/tex}.
The work done in turning the dipole through a small angle d{tex}\theta{/tex} is
dW = {tex}\tau d \theta=m B \sin \theta d \theta{/tex}
If the dipole is rotated from an initial position {tex}\theta=\theta_{1}{/tex} to the final position {tex}\theta=\theta_{2}{/tex} then the total work done will be
W = {tex}\int d W=\int_{\theta_{1}}^{\theta_{2}} m B \sin \theta d \theta=m B[-\cos \theta]_{\theta_{1}}^{\theta_{2}}{/tex}
= -mB(cos {tex}\theta_2{/tex} - cos {tex}\theta_1{/tex})
This work done is stored as the potential energy U of the dipole.
U = -mB(cos {tex}\theta_2{/tex} - cos {tex}\theta_1{/tex})
The potential energy of the dipole is zero when {tex}\vec{m} \perp \vec{B}{/tex}. So potential energy of the dipole in any orientation {tex}\theta{/tex} can be obtained by putting {tex}\theta_1{/tex} = 90° and {tex}\theta_2{/tex} = {tex}\theta{/tex} in the above equation.
U = -mB(cos {tex}\theta{/tex} - cos 90o)
or U = {tex}-m B \cos \theta=-\vec{m} \cdot \vec{B}{/tex}
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Khushi Kamboj?❤ 4 years, 6 months ago
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Amit Chauhan 4 years, 6 months ago
4Thank You