Consider the plane $\mathrm{S}$ formed by the dipole axis and the axis of earth. Let $\mathrm{P}$ be point on the magnetic equator and in $\mathrm{S}$. Let $\mathrm{Q}$ be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at $\mathrm{P}$ and $\mathrm{Q}$
Let point $\mathrm{P}$ is in the plane $\mathrm{S}$, needle is in north, so the declination is zero.
Since point $P$ lies in plane $S$ formed by the dipole axis and the axis of the Earth, declination is zero. Since point Q lies on the magnetic equator angle of dip is zero.
It is $11.3^{\circ}$ tilted with its axis of the earth so declination between $P$ and $Q$ is $11.3^{\circ}$.
The values of the apparent angles of dip in two planes at right angles to each other are $45^{\circ}$ and $30^{\circ}$ respectively. The true value of angle of dip at the place is ............
A fighter plane of length $20\, m$, wing span (distance from tip of one wing to the tip of the other wing) of $15\,m$ and height $5\,m$ is lying towards east over Delhi. Its speed is $240\, ms^{-1}$ . The earth's magnetic field over Delhi is $5 \times 10^{-5}\,T$ with the declination angle $ \sim {0^o}$ and dip of $\theta$ such that $\sin \,\theta = \frac{2}{3}$. If the voltage developed is $V_B$ between the lower and upper side of the plane and $V_W$ between the tips of the wings then $V_B$ and $V_W$ are close to
Name the elements of the earth’s magnetic field.
A magnet hung at $45^{\circ}$ with magnetic meridian makes an angle of $60^{\circ}$ with the horizontal. The actual value of the angle of dip is.
The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} Wb / m ^{2}$. If the angle of dip at that place is $60^{\circ}$, then the value of vertical component of the earth's magnetic field will be ........ $\times 10^{-4}\;W b / m^{2}$