The horizontal component of earth's magnetic field at a place is `3.5times10^(-5)T` .A very long straight conductor carrying current of `sqrt(2)A` in the direction from South east to North West is placed.The force per unit length experienced by the conductor is______________`times10^(-6)N"/"m.`

The horizontal component of earth’s magnetic field at a place is sqrt""3 times the vertical component. The angle of dip at that place is

Calculate the magnetic field at a distance of 5 cm from an infinite straight conductor carrying a current of 10 A.

Two long straight parallel conductors separated by a distance of 0.5m carry currents of 5A and 8A in the same direction. The force per unit length experienced by each other is

The horizontal component of earth's magnetic field at a certain place is 3.0 xx 10^(-5) T and having a direction from the geographic south to geographic north. The force per unit length on a very long straight conductor carrying, a 12 A in east to west direction is

The vertical component of earth's magnetic field at a place is sqrt3 times the horizontal component the value of angle of dip at this place is

Horizontal component of earth’s magnetic field at a place is sqrt()3 times the vertical component. What is the value of inclination at that place?

The horizontal component of earth's magnetic field is sqrt3 times the vertical component . What is the value of angle of dip at this place ?

The horizontal component of the earth's magnetic field is 3.6 xx 10^(-5) T where the dip is 60^@ . Find the magnitude of the earth's magnetic field.

The horizontal component of the earth's magnetic field at a certain place is 3 xx 10^-5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1 A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west, (b) south to north?