Question

When the upper tip of a magnetic dip needle is loaded with a small mass, the dip angle is found to drop from ${45}^{\circ }$ to ${30}^{\circ }$. If the total magnetic intensity at the place is $0.42$ Oersted and the pole strength of the magnetic needle is $200$ CGS units, the mass attached is $x\times {10}^{-3}\text{g}$ where $x$ is _______.

$[\sin {15}^{\circ }=0.2588;\cos {30}^{\circ }=0.8660]$

Answer 1

Before the loading of the mass, dip angle was ${45}^{\circ }$.

So, the direction of magnetic field $B$ is ${45}^{\circ }$ from horizontal.

Now, the mass is loaded, and equilibrium angle is found to be ${30}^{\circ }$.

From rotational equilibrium, torque due to weight of the mass should be equal and opposite to torque due to magnetic field.

$\therefore |\overrightarrow{r}\times \overrightarrow{F}|=|\overrightarrow{M}\times \overrightarrow{B}|$

$\Rightarrow Mg\times L\cos {30}^{\circ }=m\left(2L\right)B\sin ({45}^{\circ }-{30}^{\circ })$

$\Rightarrow M=\frac{2mB\sin {15}^{\circ }}{g\cos {30}^{\circ }}$

$\Rightarrow M=\frac{2\times 200\times 0.42\times 0.2588}{980\times 0.8660}$

$\Rightarrow M=0.051\text{g}=51\times {10}^{-3}\text{g}=x\times {10}^{-3}\text{g}$

$\therefore x=51$