Sol. (a) South pole develops at q, current induced must be clockwise at q.
In the coil, induced current is from p to q.
(b) Coil p q in the would develop S – pole at q and coli XY would also develop S pole at X. Therefore, induced current in coli p q will be from q and induced current in coil XY will be from X to Y.
(c) Induced current in the right loop will be along XYZ.
(d) Induced current in the loft loop will be along ZYX as seen from front.
(e) Induced current in the right coil is from x to Y.
(f) No current is indued because magnetic lines of force lie in the plane of the loop.
Q.2. Use Lenz’s law to determine the direction of induced current in the situation described
(a) a wire of irregular shape turning into a circular shape (b) a circular loop being deformed into a narrow straight were. The crosses indicate the magnetic field into the paper and the dots indicate magnetic field out of the paper.
Sol. (a) When a wire of irregular shape turns into a circular loop, area of the loop tends to increase. Therefore, magnetic flux linked with the loop increases. According to Lenz’s law, the direction of induced current must oppose the magnetic field, for which induced current should flow along
(b) In this case, the magnetic flux tends to decrease. Therefore, induced current must support the magnetic field, for which induced current should flow along adcba.
Q.7. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5-0 ms-1 at right angles to the horizontal component of the earth’s magnetic field 0.30×10-4 Wb m2.
(a) What is the instantaneous value of the e. m. f. induced in the were?
(b) What is the direction of e. m. f. (c) Which end of the wire is at higher electric potential?
(b) According to Fleming’s right hand rule, the direction of induced e. m. f. is from west to east.(c)West end of the wire must be at higher electric potential.
The source of this power is the external agency changing the magnetic field with time.
The direction of induced current is such as to increase the flux through the loop along positive Z – direction. Fore example, for the observer, if the loop moves to the right, the current will be seen to be anticlockwise.
Now, initial flux/turn when coil is normal to the field,
Q.14. shows a metal rod PQ resting on the rails A, B and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutually perpendicular directions. A galvanometer connects the ails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of closed loop containing the rod = 9.0 m . Answer the followi0 ng questions :
(a) Suppose K is open and the rod moves with a speed of 12 cm/s in the direction shown. Give the polarity and magnitude of induced e. m. f.
(b) Is there an excess chare 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 electron 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 ?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm/s) when K is closed.
(f) How much power is dissipated as heat in the closed circuit ? What is the source of this power ?
(g) What is the induced e. m. f. in the moving rod when the permanent magnet is rotated to a vertical position so that the field is parallel to the rails instead of being perpendicular ?
According to Fleming’s left hand rule, the direction of Lorentz force on electrons in PQ is from P to Q.
Therefore, P would acquire positive charge/ polarity & Q would acquire negative charge/polarity.
(b) Yes, excess positive charge developes at P and an equal negative excess charge developes at Q, when K is open.
(c) This is because the presence of excess charge at the ends P and Q sets up an electric field such that force due to electric field
Q.15. An air cored solenoid with length 30 cm, area of cross- section 25 cm2 and number of turns 500 carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10-3 s. How much is the average back e. m. f. induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Q.16. (Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in
(b) Evaluate the induced emf in the pool if the wire carries a current of 50 A and the loop has an instan-taneous velocity v = 10 ms-1 at thelocation x = 0.2 m, as shown. Take a = 0.1 m and assume that the loop has a large resistance.
Sol. (a) Consider a strip of width dx (of the square loop) at a distance x field due to current carrying were carrying current.
Magnetic field due to current carrying wire at a distance x