
A current carrying coil is placed with its axis perpendicular to N-S direction. Let horizontal component of earth’s magnetic field be ${H_0}$ and magnetic field inside the loop be H. If a magnet is suspended inside the loop, it makes angle angle $\theta $ with H. Then $\theta $=
1. ${\tan ^{ - 1}}\left( {\dfrac{{{H_0}}}{H}} \right)$
2. ${\tan ^{ - 1}}\left( {\dfrac{H}{{{H_0}}}} \right)$
3. $\cos e{c^{ - 1}}\left( {\dfrac{H}{{{H_0}}}} \right)$
4. ${\cot ^{ - 1}}\left( {\dfrac{{{H_0}}}{H}} \right)$
Answer
591.6k+ views
Hint: In the solution we will use the concept of magnetic field direction in a magnet and the horizontal and vertical component of earth’s magnetic field and magnetic field inside the loop.
Complete step by step answer:
Given:
The coil is placed with its axis perpendicular to N-S direction.
The horizontal component of earth’s magnetic field is ${H_0}$.
The magnetic field inside the loop is $H$ .
As the coil is placed perpendicular to N-S direction means the coil would be placed along the E-W direction axis. So the horizontal component of earth’s magnetic field would be acting in the north direction. The magnetic field inside the loop depends upon the direction of the current in the coil. So let's consider the direction of the magnetic field inside the loop is in the east direction. As the magnet is suspended inside the loop it makes an angle $\theta $ from the axis of magnetic inside the loop (horizontal positive $x$-axis). So the angle $\theta $ can be calculated as,
$
\tan \theta = \dfrac{{{H_0}}}{H}\\
\therefore\theta = {\tan ^{ - 1}}\left( {\dfrac{{{H_0}}}{H}} \right)
$
Therefore, the option (1) is the correct answer that is $\theta = {\tan ^{ - 1}}\left( {\dfrac{{{H_0}}}{H}} \right)$.
Note: Remember the magnetic field inside the loop $H$ depends upon the direction of current inside the coil. Between the horizontal components of earth magnetic field and the magnetic field of the coil, the suspended magnet makes an angle $\theta $ because when a magnet is placed in a magnetic field then a torque acts on the magnet along the resultant magnetic field. The dipole moment of the magnet always acts from south to North Pole of the magnet.
Complete step by step answer:
Given:
The coil is placed with its axis perpendicular to N-S direction.
The horizontal component of earth’s magnetic field is ${H_0}$.
The magnetic field inside the loop is $H$ .
As the coil is placed perpendicular to N-S direction means the coil would be placed along the E-W direction axis. So the horizontal component of earth’s magnetic field would be acting in the north direction. The magnetic field inside the loop depends upon the direction of the current in the coil. So let's consider the direction of the magnetic field inside the loop is in the east direction. As the magnet is suspended inside the loop it makes an angle $\theta $ from the axis of magnetic inside the loop (horizontal positive $x$-axis). So the angle $\theta $ can be calculated as,
$
\tan \theta = \dfrac{{{H_0}}}{H}\\
\therefore\theta = {\tan ^{ - 1}}\left( {\dfrac{{{H_0}}}{H}} \right)
$
Therefore, the option (1) is the correct answer that is $\theta = {\tan ^{ - 1}}\left( {\dfrac{{{H_0}}}{H}} \right)$.
Note: Remember the magnetic field inside the loop $H$ depends upon the direction of current inside the coil. Between the horizontal components of earth magnetic field and the magnetic field of the coil, the suspended magnet makes an angle $\theta $ because when a magnet is placed in a magnetic field then a torque acts on the magnet along the resultant magnetic field. The dipole moment of the magnet always acts from south to North Pole of the magnet.
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