Dip angle are $30^o$ and $45^o$ at two different places, then ratio of horizontal component of earth magnetic field at these place will be
$\sqrt 3:\sqrt 2$
$1 :\sqrt 2$
$1 :\sqrt 3$
$1 :2$
At a place, the magnitudes of the horizontal component and total intensity of the magnetic field of the earth are $0.3 $ and $ 0.6$ Oersted respectively. The value of the angle of dip at this place will be.....$^o$
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.....$G$
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
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of $30$ turns and radius $12 \;cm .$ The coil is in a vertical plane making an angle of $45^{\circ}$ with the magnetic meridian. When the current in the coil is $0.35 \;A$, the needle points west to east.
$(a)$ Determine the horizontal component of the earth's magnetic field at the location.
$(b)$ The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of $90^{\circ}$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
At a certain place the horizontal component of the earth’s magnetic field is $B_0$ and the angle of dip is $45^o$. The total intensity of the field at that place will be