The horizontal and vertical components of earth's field at a place are `0.24xx10^(-4)` and `0.40xx10^(-4)T` respectively. Calculate the angel of dip and resultant intensity of earth's field

At a place , the horizontal and vertical components of earth's magnetic field are 0.3 Oe and 0.2 Oe , respectively . Determine the resultant intensity and angle of dip there .

The vertical component of earth’s magnetic field is zero at a place when angle of dip is—

The horizontal and vertical comporient-of earth’s magnetic field at a plane are same. What is the angle of dip at that location?

The angle pf dip and the horizontal component of earth's magnetic field at a place are 30^(@)S and 0.36 Oe . Determine the magnitude and direction of the vertical component of earth 's magnetic field at thet place .

The vertical component of earth's magnetic field at a place is sqrt(3) Times of the horizontal component. What is the angle of dip at this place ?

Angle of declination at a place is 20^(@) What will be the horizontal and vertical components of earth's magnetic field on the geographical meridian at that place? Given H=2xx10^(-5)T and angle of dip at that place 60^(@)

The horizontal component of earth's magnetic field at a place is sqrt(3) Time the vertical component. What is the angle of dip at this place?

The geomagnetic field and the angle of dip at a place are 2xx10^(-5)T And 30^(@) Respectively . What are the horizontal and vertical fields at this place ?

If the horizontal component of earth's magnetic field intensity at a place is half the total intensity , determine the angle of dip there.

A proton obliquely enters a uniform magnetic field of 0.5 T. It the component of its velocity along and perpendicular to the direction of the magnetic field are 1.5xx10^(5)m.s^(-1) and 2xx10^(5)m.s^(-1) , respectively, calculate the radius of the helical path followed by the proton ant its pitch. [m=1.67xx10^(-27)kg,e=1.6xx10^(-19)C]