Dip angle in vertical plane at an angle | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\alpha=\cos ^{-1} \frac{1}{\sqrt{2}}, \Rightarrow \cos \alpha=\frac{1}{\sqrt{2}}, 	heta^{\prime}=60^{\circ}$

$	an 	heta^{\prime}=	an 	heta /

Vedclass · Accepted Answer

${	an ^{ - 1}}\,\left( {\sqrt {\frac{3}{2}} } ight)$

Vedclass · Answer

$\alpha=\cos ^{-1} \frac{1}{\sqrt{2}}, \Rightarrow \cos \alpha=\frac{1}{\sqrt{2}}, 	heta^{\prime}=60^{\circ}$

$	an 	heta^{\prime}=	an 	heta / \cos \alpha$

so $	an 	heta=	an 	heta^{\prime} \cos \alpha$

$=\left(	an 60^{\circ}ight) \frac{1}{\sqrt{2}}=\sqrt{3} 	imes \frac{1}{\sqrt{2}}$

$	heta=	an ^{-1}(\sqrt{\frac{3}{2}})$

Dip angle in vertical plane at an angle ${\cos ^{ - 1}}\,\left( {\frac{1}{{\sqrt 2 }}} \right)$ from magnetic meridian is $60^o$, then actual dip at that place

Similar Questions

At magnetic poles of earth, angle of dip is.....$^o$

If $\theta _1$ and $\theta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\theta$ is given by

The magnetic needle of a tangent galvanometer is deflected at an angle $30^o$ due to a magnet. The horizontal component of earth’s magnetic field $0.34 \times {10^{ - 4}}\,T$ is along the plane of the coil. The magnetic intensity is

Choose the correct option:

The angle of dip at a certain place on earth is $60^o$ and the magnitude of earth's horizontal component of magnetic field is $0.26\, G$. The magnetic field at the place on earth is.....$G$