The magnetic moment of a current-carrying loop is $2.1 \times 10^{-25} \text{ A m}^2$. The magnetic field at a point on its axis at a distance of $1 \text{ Å}$ is:

$A$ magnet of magnetic moment $2 \,J \,T^{-1}$ is aligned in the direction of a magnetic field of $0.1 \,T$. What is the net work done to bring the magnet normal to the magnetic field (in $\,J$)?

$A$ bar magnet placed in a uniform magnetic field making an angle $\theta$ with the field experiences a torque. If the angle made by the magnet with the field is doubled,the torque experienced by the magnet increases by $41.4 \%$. The initial angle made by the magnet with the magnetic field is (in $^{\circ}$)

Two identical magnetic dipoles of magnetic moments $1.0 \, A-m^2$ each are placed at a separation of $2 \, m$ with their axes perpendicular to each other. The resultant magnetic field at a point $P$ midway between the dipoles is:

$A$ short bar magnet of magnetic moment $0.21 \ A \cdot m^2$ is placed with its axis perpendicular to the direction of the horizontal component of the earth's magnetic field. The distance of the point on the axis of the magnet from the centre of the magnet where the resultant magnetic field is inclined at $45^{\circ}$ with the horizontal component of the earth's field direction is (horizontal component of the earth's magnetic field $= 4.2 \times 10^{-5} \ T$). (in $cm$)

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 mid-point $P$,at an angle $\theta = 45^\circ$ with the horizontal line as shown in the figure. What would be the magnitude of the force on the particle at that instant? ($d$ is much larger than the dimensions of the dipole)