Two equal bar magnets are kept as shown in the figure. The direction of the resultant magnetic field,indicated by the arrowhead at point $P$,is (approximately):

Two identical bar magnets of magnetic moment $M$ each, are placed along $X$ and $Y$-axes, respectively at a distance $d$ from the origin (as shown in the figure). The origin lies on the perpendicular bisector of the magnet placed on the $X$-axis and on the magnetic axis of the magnet placed on the $Y$-axis. If the magnitude of the total magnetic field at the origin is $B = \alpha \left[ \frac{\mu_0}{4 \pi} \frac{M}{d^3} \right]$, then the value of the constant $\alpha$ will be (given $d >> l$, where $l$ is the length of the bar magnets and the direction of $N$ to $S$ in the magnets is opposite with respect to each other).

$A$ short magnetic needle is placed in a magnetic field $B \hat{i}$. The needle is oriented in the direction $(\sqrt{3} \hat{i} + \hat{j})$. The needle experiences a torque of $0.06 \ N-m$. If the same magnetic needle is placed in a magnetic field $2B \hat{j}$ and is oriented in the direction $(\hat{i} + \sqrt{3} \hat{j})$,the torque experienced by it is: (in $N-m$)

$A$ magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^{\circ},$ and one of the fields has a magnitude of $1.2 \times 10^{-2}\;T$. If the dipole comes to stable equilibrium at an angle of $15^{\circ}$ with this field,what is the magnitude of the other field?

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

What is the magnitude of the equatorial magnetic field due to a bar magnet of length $5.0 \ cm$ at a distance $75 \ cm$ from its midpoint? The magnetic moment of the bar magnet is $0.75 \ A \ m^2$.