A copper wire of uniform area of cross-section 3.4×10−5 m23.4×10^{-5}\space\text m^23.4×10−5 m2 carries a current of 4.0 A. The magnitude of the electric field applied is ____________ (Resistivity of copper: 1.7×10−81.7×10^{-8}1.7×10−8 Ω m).

2.0×10−3 Vm2.0×10^{-3}\space$$\frac{\text V}{\text m}2$$.0×10−3 mV

1.0×10−3 Vm1.0×10^{-3}\space$$\frac{\text V}{\text m}1$$.0×10−3 mV

3.4×10−2 Vm3.4×10^{-2}\space$$\frac{\text V}{\text m}3$$.4×10−2 mV

1.6×10−2 Vm1.6×10^{-2}\space$$\frac{\text V}{\text m}1$$.6×10−2 mV

A copper wire of uniform area of cross-section 3.4×10−5 m23.4×10^{-5}\space\text m^23.4×10−5 m2 carries a current of 4.0 A. The magnitude of the electric field applied is ____________ (Resistivity of copper: 1.7×10−81.7×10^{-8}1.7×10−8 Ω m).

Given:Area of cross-section, A=3.4×10−5 m2Current, I=4.0 AResistivity of copper, ρ=1.7×10−8 Ω⋅m$$\\textbf{Given:}$$ \\\\$$\\begin{aligned}$$&$$\\text{Area of cross-section, }$$ A = 3.4 \\times 10^{-5} \\, $$\\text{m}$$^2 \\\\&$$\\text{Current, }$$ I = 4.0 \\, $$\\text{A}$$ \\\\&$$\\text{Resistivity of copper, }$$ \\rho = 1.7 \\times 10^{-8} \\, \\Omega \\cdot $$\\text{m}$$$$\\end{aligned}$$Given:Area of cross-section, A=3.4×10−5m2Current, I=4.0AResistivity of copper, ρ=1.7×10−8Ω⋅mLet the length of the copper wire, L=1 m$$\\text{Let the length of the copper wire, }$$ L = 1 \\, $$\\text{m}$$Let the length of the copper wire, L=1mV=IR=I⋅ρ⋅LAV=4⋅1.7×10−8⋅13.4×10−5V=4⋅1.7×10−83.4×10−5=2.0×10−3 VV = IR = I \\cdot \\rho \\cdot $$\\frac{L}{A}$$ \\\\[8pt]V = 4 \\cdot 1.7 \\times 10^{-8} \\cdot $$\\frac{1}$${3.4 \\times 10^{-5}} \\\\[8pt]V = \\frac{4 \\cdot 1.7 \\times 10^{-8}}{3.4 \\times 10^{-5}} = 2.0 \\times 10^{-3} \\, $$\\text{V}$$V=IR=I⋅ρ⋅ALV=4⋅1.7×10−8⋅3.4×10−51V=3.4×10−54⋅1.7×10−8=2.0×10−3VE=VL=2.0×10−31=2.0×10−3 V/mE = $$\\frac{V}{L}$$ = \\frac{2.0 \\times 10^{-3}}{1} = 2.0 \\times 10^{-3} \\, $$\\text{V/m}$$E=LV=12.0×10−3=2.0×10−3V/m

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A copper wire of uniform area of cross-section 3.4×10−5 m23.4×10^{-5}\space\text m^23.4×10−5 m2 carries a current of 4.0 A. The magnit

Question

A copper wire of uniform area of cross-section $3.4×10^{-5}\space\text m^2$ carries a current of 4.0 A. The magnitude of the electric field applied is ____________ (Resistivity of copper: $1.7×10^{-8}$ Ω m).

$2.0×10^{-3}\space\frac{\text V}{\text m}$

$1.0×10^{-3}\space\frac{\text V}{\text m}$

$3.4×10^{-2}\space\frac{\text V}{\text m}$

$1.6×10^{-2}\space\frac{\text V}{\text m}$

Solution

Correct option is A

$\textbf{Given:} \\\begin{aligned}&\text{Area of cross-section, } A = 3.4 \times 10^{-5} \, \text{m}^2 \\&\text{Current, } I = 4.0 \, \text{A} \\&\text{Resistivity of copper, } \rho = 1.7 \times 10^{-8} \, \Omega \cdot \text{m}\end{aligned}$

$\text{Let the length of the copper wire, } L = 1 \, \text{m}$

$V = IR = I \cdot \rho \cdot \frac{L}{A} \\[8pt]V = 4 \cdot 1.7 \times 10^{-8} \cdot \frac{1}{3.4 \times 10^{-5}} \\[8pt]V = \frac{4 \cdot 1.7 \times 10^{-8}}{3.4 \times 10^{-5}} = 2.0 \times 10^{-3} \, \text{V}$

$E = \frac{V}{L} = \frac{2.0 \times 10^{-3}}{1} = 2.0 \times 10^{-3} \, \text{V/m}$

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A copper wire of uniform area of cross-section $3.4×10^{-5}\space\text m^2$ carries a current of 4.0 A. The magnitude of the electric field applied is ____________ (Resistivity of copper: $1.7×10^{-8}$ Ω m).

$2.0×10^{-3}\space\frac{\text V}{\text m}$

$1.0×10^{-3}\space\frac{\text V}{\text m}$

$3.4×10^{-2}\space\frac{\text V}{\text m}$

$1.6×10^{-2}\space\frac{\text V}{\text m}$

Correct option is A

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A copper wire of uniform area of cross-section 3.4×10−5 m23.4×10^{-5}\space\text m^23.4×10−5 m2​ carries a current of 4.0 A. The magnitude of the electric field applied is ____________ (Resistivity of copper: 1.7×10−81.7×10^{-8}1.7×10−8​ Ω m).

​2.0×10−3 Vm2.0×10^{-3}\space\frac{\text V}{\text m}2.0×10−3 mV​​​

​1.0×10−3 Vm1.0×10^{-3}\space\frac{\text V}{\text m}1.0×10−3 mV​​​

​3.4×10−2 Vm3.4×10^{-2}\space\frac{\text V}{\text m}3.4×10−2 mV​​​

​1.6×10−2 Vm1.6×10^{-2}\space\frac{\text V}{\text m}1.6×10−2 mV​​​

Correct option is A

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Access ‘AAI JE ATC’ Mock Tests with

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A copper wire of uniform area of cross-section $3.4×10^{-5}\space\text m^2$ carries a current of 4.0 A. The magnitude of the electric field applied is ____________ (Resistivity of copper: $1.7×10^{-8}$ Ω m).

$2.0×10^{-3}\space\frac{\text V}{\text m}$

$1.0×10^{-3}\space\frac{\text V}{\text m}$

$3.4×10^{-2}\space\frac{\text V}{\text m}$

$1.6×10^{-2}\space\frac{\text V}{\text m}$