Question 236: Current Electricity

Question

A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m2. At every point the direction of current density is at an angle of 60$$^\circ$$ with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is :(Resistivity of magnesium $$ho$$ = 44 $$	imes$$ 10$$-$$8 $$\Omega$$m)

Accepted Answer

Given, current, I = 5A
Area of cross-section of wire, A = 0.04 m
2
We know that, $$J = {I \over A}$$
$$ \Rightarrow I = JA$$
or $$I = J\,.\,A$$ or $$I = JA\cos 	heta $$
where, J = current density.
$$ \Rightarrow 5 = J\left( {{4 \over {100}}} ight) 	imes \cos (60^\circ )$$ [$$\because$$ Given, $$	heta$$ = 60$$^\circ$$]
$$J = 500 	imes {1 \over 2}$$ [$$\because$$ cos60$$^\circ$$ = $${1 \over 2}$$]
$$\Rightarrow$$ J = 250 Am
$$-$$2
The relation between electric field, current density and resistivity can be given as,
E = $$ho$$ . J
= 44 $$	imes$$ 10
$$-$$8
$$	imes$$ 250 [$$\because$$ Resistivity, $$ho$$ = 44 $$	imes$$ 10
$$-$$8
$$\Omega$$-m]
= 11 $$	imes$$ 10
$$-$$5
V/m

Answer

11 $$	imes$$ 10$$-$$5 V/m

Answer

11 $$	imes$$ 10$$-$$3 V/m

Answer

11 $$	imes$$ 10$$-$$7 V/m

Answer

11 $$	imes$$ 10$$-$$2 V/m

Current Electricity

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