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silvercameron:
thankyou sir. But i dont understand that how could you conclude > "diameter of copper=(1.7*10^-8)/(2.6*10^-8)*1.2mm

I tried this today and did it like this:
resistance of aluminium = resistance of copper.

So :    (rho*l)/A  of aluminium= (rho*l)/ A of copper.

 Radius of aluminium = 1.2/2 = 0.6 mm = 6*10-4 m
Area of aluminium= pi* r2 = pi* (6*10-4)2 = 1.131*10-6

therfore :     ((2.6*10-8)*l)/(1.13*10-6)=((1.7*10-8)*l)/pi*r2

l and l gets canceled cos they are the same.
(2.6*10-8)/(1.131*10-6)= (1.7*10-8)/(pi*r2)
we cross-multiply, get the radius, multiply it by 2 and finally get diameter= 9.72*10 -4 m

^^is my method correct?

astarmathsandphysics:
Yes. I should have square rooted.

ashwinkandel:
Questions from Electromagnetism
1.
Figure shows a wire XY which carries a direct current. Plotting compass R, placed alongside the wire, points due north. Compass P is placed below the wire and compass Q is placed above the wire.
a. State the direction of the current in the wire.
b. State in which direction compass P points
c. State in wihch direction compass Q points if the current in the wire is reversed.

2.
At a point on the Earth's surface the horizontal component of the Earth's magnetic field is 1.6*10^-5 T. A piece of wire 3.0 m long and of weight 0.020N lies in the east-west direction on a labortary bench. When a large current flows in the wire, the wire just lifts off the surface of the bench.
i. State the direction of the current in the wire.[1]
ii. Calculate the minimum current needed to lifet the wire from the bench. [3]

astarmathsandphysics:
a.righthand grip rule current flows from Y yo X
b. same as Q
c.North West
2. by flemings left hand rule current is from west to east
ii. F=BIl
0.02=1.6*10^-5*I*3 so I=0.02/(1.6*10^-5*3)417A

ashwinkandel:
Can you tell me actually the use of the compass here. What does it indicate or shows?

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