The magnetic field at center O of the arc in figure is :

A straight section PQ of a circuit lies along the X-axis form x = –a/2 to x = a/2 and carries a steady current i. The magnetic field due to the section PQ at a point X = +a will be :

A long straight wire carrying a current of 30A is placed in an external uniform magnetic field
of induction \[4\times 10^{-4} T\]. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point 2.0 cm away from the wire is
\[\left( \mu_{0}=4\pi\times 10^{-7}H/m \right)\]:

Two parallel long wires carry currents i1 and i2 with i1 > i2. When the currents are in the same
direction, the magnetic field midway between the wires is 10μT. When the direction of i2 is reversed, it becomes 40μT. the ratio i1/i2 is :

The magnetic field at the centre of an equilateral triangular loop of side 2L and carrying a current i is :

Two wires AO and OC carry equal currents i as shown in figure. One end of both the wire extends to infinity. Angle AOC is α. The magnitude of magnetic field at a point P on the bisector of these two wires at a distance r from point O is :

In the above figure magnetic field at point C will be:

A ring of radius r is carrying a current I. The magnetic field B is always perpendicular to the ring as shown in Fig. The force on the ring is :

A wire carrying a current i is placed in a uniform magnetic field in the form of the curve
y =a sin (πx/L), 0 ≤ x ≤ 2L. The force acting on the wire is :

Two thick wires and two thin wires, all of the same materials and same length form a square in
the three different ways P, Q and R as shown in fig with current connection shown, the magnetic field at the centre of the square is zero in cases: