Physics, asked by tripathiom346, 27 days ago

The moment of force about a fixed end P of a rod PQ of length 480 cm is 3.36 Nm. What is the force applied at the free end Q of the rod?​

Answers

Answered by shifaalam
0

Explanation:

Solution

verified

Verified by Toppr

(a)

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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