Field of a Magnet & Magnet in a Uniform Field

Two identical small bar magnets each of dipole moment $3\sqrt{5}\text{ J/T}$ are placed at a center to center separation of 10 cm, with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is $\alpha \times 10^{-3}\text{ T}$. The value of $\alpha$ is $\underline{\qquad}\ \left(\mu_0 = 4\pi \times 10^{-7}\text{ Tm/A}\right)$

A magnetic dipole lying parallel to a magnetic field requires '$W$' units of work to turn it through $30^\circ$. The torque required to keep this needle in this position is

$\left[\sin 30^\circ = \cos 60^\circ = 0.5, \sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 90^\circ = \cos 0^\circ = 1\right]$

A short bar magnet placed with its axis at $30^\circ$ with an external field of 800 Gauss, experiences a torque of $0.016\text{ N.m}$. The work done in moving it from most stable to most unstable position is $\alpha \times 10^{-3}\text{ J}$. The value of $\alpha$ is ______.

The work done in turning a magnet of magnetic moment ' $M$ ' by an angle of $90^\circ$ from the meridian is ' $n$ ' times the corresponding Work done to turn it through an angle of $60^\circ$ where the value of ' $n$ ' is $(\cos 90^\circ = 0, \cos 60^\circ = 0.5)$

A magnetic dipole experiences a torque of $80\sqrt{3}\text{ N m}$ when placed in uniform magnetic field in such a way that dipole moment makes angle of $60^\circ$ with magnetic field. The potential energy of the dipole is:

A bar magnet has total length $2l = 20$ units and the field point P is at a distance $d = 10$ units from the centre of the magnet. If the relative uncertainty of length measurement is $1\%$, then uncertainty of the magnetic field at point P is:

The magnetic potential energy, when a magnetic bar of magnetic moment $\vec{m}$ is placed perpendicular to the magnetic field $\vec{B}$ is

The magnetic moment of a bar magnet is $0.5\text{ Am}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2}\text{ T}$. The work done in rotating it from its most stable to most unstable position is:

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20\text{ cm}$ from its center is $1.5 \times 10^{-5}\text{ T m}$. The magnetic moment of the dipole is _________ $\text{A m}^2$. (Given: $\frac{\mu_0}{4\pi} = 10^{-7}\text{ TmA}^{-1}$)

A bar magnet with a magnetic moment $5.0\text{ Am}^2$ is placed in parallel position relative to a magnetic field of $0.4\text{ T}$. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is _____________.

A bar magnet having a magnetic moment of 2.0 $\times$ 105 JT${}^{-1}$, is placed along the direction of uniform magnetic field of magnitude B = 14 $\times$ 10$-$5 T. The work done in rotating the magnet slowly through 60${}^\circ$ from the direction of field is:

A magnetic dipole of magnetic moment $M$, is freely suspended in a magnetic field of induction B. The minimum and maximum values of potential energy of the dipole, respectively are

Two short magnetic dipoles m1 and m2 each having magnetic moment of 1 Am2 are placed at point O and P respectively. The distance between OP is 1 meter. The torque experienced by the magnetic dipole m2 due to the presence of m1 is ........... $\times$ 10$-$7 Nm.

Two identical bar magnets each of magnetic moment $M$, separated by some distance are kept perpendicular to each other. The magnetic induction at a point at the same distance $d$ from the centre of magnets, is ($\mu_0 =$ permeability of free space)

A small bar magnet placed with its axis

at 30° with an external field of 0.06 T

experiences a torque of 0.018 Nm. The

minimum work required to rotate it from its

stable to unstable equilibrium position is:

Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing, through their midpoint P, at angle $\theta$ = 45° with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant?

(d is much larger than the dimensions of the dipole)

A magnet of total magnetic moment 10−2 $\hat{i}$ A-m2 is placed in a time varying magnetic field, B$\hat{i}$ (cos $\omega t$) where B = 1 Tesla and $\omega$ = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is -

A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 $\mu$A and subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is

A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 weber/m2. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

The mid points of two small magnetic dipoles of length d in end-on positions, are separated by a distance x, (x > > d). The force between them is proportional to x$-$n where n is:

A bar magnet of magnetic moment M is placed at right angles to a magnetic induction B. If a force F is experienced by each pole of the magnet, the length of the magnet will be

A current loop in a magnetic field

Two short bar magnets of length $1\text{ cm}$ each have magnetic moments $1.20\text{ Am}^{2}$ and $1.00\text{ Am}^{2}$ respectively. They are placed on a horizontal table parallel to each other with their $N$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $20.0\text{ cm.}$ The value of the resultant horizontal magnetic induction at the mid-point $O$ of the line joining their centres is close to $($ Horizontal component of earth's magnetic induction is $3.6 \times 10^{-5}\text{ Wb/m}^{2})$

A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}J$ of work to turn it through 60°. The torque needed to maintain the needle in this position will be

A short bar magnet of magnetic moment 0.4 J T${}^{-1}$ is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is

Two identical bar magnets are fixed with their centers at a distance d apart. A stationary charge Q is placed at P in between the gap of the two magnets at a distance D from the centre O as shown in the figure

The force on the charge Q is

A bar magnet having a magnetic moment of 2 $\times$ 104 J T${}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 $\times$ 10$-$4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is

A magnetic needle is kept in a non-uniform magnetic field. It experiences:

A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $\vec{B}$is in plane of the coil. If due to a current i in the triangle a torque $\tau$ acts on it, the side $l$ of the triangle is

A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^\circ.$ The torque needed to maintain the needle in this position will be:

A bar magnet of magnetic moment $\vec{M}$ is placed in a magnetic field of induction $\vec{B}$. The torque exerted on it is

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^\circ$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^\circ$. The value of $n$ is given by