1. A 1-meter long aluminum cylindrical wire of 2 mm^2 cross-sectional area has a
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Question
1. A 1-meter long aluminum cylindrical wire of 2 mm^2 cross-sectional area has a uniform current of 10 amperes in it. The resistivity of aluminum is 2.63 * 10^-8 ohm-meter. The electric field along the wire would be:
It should be
a)1.9 *10^-6 n/C
b)1.32*10^-1n/C.
Please show work. Thanks
2. An inductance-capacitance (LC) circuit in 1, for an oscillating mass m on the spring (constant k) in 2 and for a simple pendulum of length l in 3, the quantities that have the common unit of "Hertz" are:
It should be
a) sqrt(LC), sqrt(m/k), sqrt(g/l)
b)sqrt(1/LC), sqrt(k/m), sqrt(g/l)
c)sqrt(1/LC), sqrt(m/k), sqrt(l/g)
Please show work. Thanks
Explanation / Answer
here,
length of wire , l = 1 m
cross sectional area = 2 mm^2 = 2 * 10^-6 m^2
current , I = 10 A
resistivity , p = 2.63 * 10^-8 ohm.m
the potential difference , V = R * I
V = p * l /area * I
the electric feild , E = V/l
E = p * I /area
E = 2.63 * 10^-8 * 10 /( 2 * 10^-6) V/m
E = 1.32 * 10^-1 N/C
the electric feild is b) 1.32 * 10^-1 N/C
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