$A$ 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 $N-m$)

$A$ 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 $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position 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 $0.75 \ A \ m^2$.

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

Two short bar magnets $A$ and $B$ (having magnetic moments $M_{1}$ and $M_{2}$ respectively) are kept one above the other with their magnetic axes perpendicular to each other. If their resultant magnetic field at a point on the axis of magnet $A$ is inclined at $45^{\circ}$ with the axis of magnet $A$,then the ratio of magnetic moments $\frac{M_{2}}{M_{1}}$ is $[\tan 45^{\circ} = 1]$.

Two short magnets $AB$ and $CD$ are in the $X-Y$ plane and are parallel to the $X$-axis. The coordinates of their centres are $(0,2)$ and $(2,0)$ respectively. The line joining the north-south poles of $CD$ is opposite to that of $AB$ and lies along the positive $X$-axis. The resultant magnetic field induction due to $AB$ and $CD$ at a point $P(2,2)$ is $100 \times 10^{-7} \ T$. When the poles of the magnet $CD$ are reversed, the resultant field induction is $50 \times 10^{-7} \ T$. The values of the magnetic moments of $AB$ and $CD$ (in $Am^2$) are: