90 THE WORLD OF MATHEMATICS VT 4. There are markings as shown on a domestic appliance. Through how many de does the pointer turn when it moves (1) (a) clockwise from off to cold? (b) anti-clockwise from hot to warm? (c) clockwise from very hot to cool? (d) clockwise from cold to warm? (e) anti-clockwise from very cold to very hot? clockwise from warm to cool? (ii) Which setting is the pointer facing, after it turns (a) 45° clockwise from normal ? (b) 135° anti-clockwise from off? (c) 360° anti-clockwise from hot ? (d) 270° clockwise from cold? . (i) Through what angle does the minute-hand of a clock turn in 45 minutes the hour-hand in 30 minutes ? (ii) What rotation is needed to turn direction?
Answers
Answer:
Induce EMF is e=Blv=9×10
−3
V
The polarity of the induced emf is such that end P shows positive while end Q shows negative ends.
(b)
Yes; when key K is closed, excess charge is maintained by the continuous flow of current.
When key K is open, there is excess charge built up at both ends of the rods.
When key K is closed, excess charge is maintained by the continuous flow of current.
(c)
Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.
(d)
Retarding force exerted on the rod, F=IBl
I=e/R=9×10
−3
/9×10
−3
=1A
so, F=1×0.5×0.15=75×10
−3
N
(e)
Retarding force, F=BIl
where, I=e/R=1A
so, F=75×10
−3
N
Also, Power P=Fv=75×10
−3
×0.12=9mW
(f)
Power dissipated = I
2
R=9mW
The source of this power external agent.
(g)
Zero. As, in this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.
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The conducting rod ab, as shown in figure makes contact with metal rails ca and db. The apparatus is in a uniform magnetic field of 0.800T, perpendicular to the plane of the figure. If the resistance of the circuit abdc is 1.50Ω (assumed to be constant), find the force (magnitude and direction) required to keep the rod moving to the right with a constant speed of 7.50m/s. You can ignore friction.
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Electric charge q is distributed uniformly over a rod of length l. The rod is placed parallel to a long wire carrying a current i. The separation between the rod and the wire is a. The force needed to move the rod along its length with a uniform velocity v is :
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Answer:
Induce EMF is e=Blv=9×10
−3
V
The polarity of the induced emf is such that end P shows positive while end Q shows negative ends.
(b)
Yes; when key K is closed, excess charge is maintained by the continuous flow of current.
When key K is open, there is excess charge built up at both ends of the rods.
When key K is closed, excess charge is maintained by the continuous flow of current.
(c)
Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.
(d)
Retarding force exerted on the rod, F=IBl
I=e/R=9×10
−3
/9×10
−3
=1A
so, F=1×0.5×0.15=75×10
−3
N
(e)
Retarding force, F=BIl
where, I=e/R=1A
so, F=75×10
−3
N
Also, Power P=Fv=75×10
−3
×0.12=9mW
(f)
Power dissipated = I
2
R=9mW
The source of this power external agent.
(g)
Zero. As, in this case, no emf is induced in the coil because the motion of the rod does not cut across the