Figure shows a part of an electric circuit. The wires AB, CD, and EF are long and have identical resistances. The separation between the neighbouring wires is 1.0 cm. The wires AE and BF have negligible resistances and the ammeter reads 30 A . Calculate the magnetic force per unit length of AB and CD. `

A magnetic field vecB=B_0 sin (omegat)hatk covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d as in figure. The wires are in the xy-plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity ?

The potentiometer wire of length 100cm has a resistance of 10Omega . It is connected in series with a resistance of 5Omega and an accumulaot or emf 3V having negligible resistance. A source 1.2V is balanced against a length 'L' of the potentionmeter wire. find the value of L.

Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. vecB is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. vecB is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

An electric dipole is prepared by taking two electric charges of 2 xx 10^(-8) C separated by distance 2 mm. This dipole is kept near a line charge distribution having density 4 xx 10^(-4) C/m in such a way that the negative electric charge of the dipole is at a distance 2 cm from the wire as shown in the figure. Calculate the force acting on the dipole. [Take k=1/(4piepsilon_(0)) = 9 xx 10^(9) Nm^(2)C^(-2) ]

Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod =15 cm, B =0.50 T, resistance of the closed loop containing the rod 9.0 m Omega . Assume the field to be uniform. (a) Suppose K is open and the rod is moved with a speed of 12 "cm s"^(-1) in the direction shown. Give the polarity and magnitude of the induced emf. (b) Is there an excess charge built up at the ends of the rods when K is open ? What if K is closed ? (c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain. (d) What is the retarding force on the rod when K is closed ? How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s^(-1) ) when K is closed ? How much power is required when K is open? (f)How much power is dissipated as heat in the closed circuit ? What is the source of this power ? (g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular ?

Two wires AB & CD, each 1m length, carry a total charge of 0.2 microcoulomb each and are placed as shown in figure. The ends B & C are separated 1 m distance. Determine the value of electric intensity at the point P in the vector form.Note P is the mid point of BC.

A potentiometer circuit is setup as shown. The potential gradient across the potentiometer wire is k "volt"//cm and the ammeter present in the circuit reads. 1.0 A When two-way key is switched off. The balance point, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths l_(2) cm and l_(2) cm respectively. The magnitudes, of the resistors R and X , in ohm, are then, equal, respectively, to