A vertical straight conductor carries a current vertically upwards. A point `P` lies to the east of it at a small distance and another point `Q` lies to the west at the same distance. The magnetic field at `P` is

A vertical wire carriers a current upwards. The magnetic field at a point due north of the wire is directed

The point P(a,b) lies on the straight line 3x+2y=13 and the point Q(b,a) lies on the straight line 4x-y=-5 , then the equation of the line PQ is

The point P (a, b) lies on the straight line 3x – 2y = 13 and the point Q (b, a) lies on the straight line 4x – y = 5 then the equation of line PQ is

A straight vertical conductor carries a current. At a point 5cm due north of it, the magnetic induction is founded to be 20muT due east. The magnetic induction at a point 10cm east of its will be

A point P lies on the axis of a flat coil carrying a current. The magnetic moment of the coil is mu . The distance of P from the coil is d , which is large compared to the radius of the coil. The magnetic field at P has magnitude

A straight wire of finite length carrying current I subtends an angle of 60^(@) at point P as shown. The magnetic field at P is

A proton moves horizontally towards a vertical conductor carrying a current upwards. It will be deflected