JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 3)
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 $$\rho$$ = 44 $$\times$$ 10$$-$$8 $$\Omega$$m)
(Resistivity of magnesium $$\rho$$ = 44 $$\times$$ 10$$-$$8 $$\Omega$$m)
11 $$\times$$ 10$$-$$5 V/m
11 $$\times$$ 10$$-$$3 V/m
11 $$\times$$ 10$$-$$7 V/m
11 $$\times$$ 10$$-$$2 V/m
Explanation
Given, current, I = 5A
Area of cross-section of wire, A = 0.04 m2
We know that, $$J = {I \over A}$$
$$ \Rightarrow I = JA$$
or $$I = J\,.\,A$$ or $$I = JA\cos \theta $$
where, J = current density.
$$ \Rightarrow 5 = J\left( {{4 \over {100}}} \right) \times \cos (60^\circ )$$ [$$\because$$ Given, $$\theta$$ = 60$$^\circ$$]
$$J = 500 \times {1 \over 2}$$ [$$\because$$ cos60$$^\circ$$ = $${1 \over 2}$$]
$$\Rightarrow$$ J = 250 Am$$-$$2
The relation between electric field, current density and resistivity can be given as,
E = $$\rho$$ . J
= 44 $$\times$$ 10$$-$$8 $$\times$$ 250 [$$\because$$ Resistivity, $$\rho$$ = 44 $$\times$$ 10$$-$$8 $$\Omega$$-m]
= 11 $$\times$$ 10$$-$$5 V/m
Area of cross-section of wire, A = 0.04 m2
We know that, $$J = {I \over A}$$
$$ \Rightarrow I = JA$$
or $$I = J\,.\,A$$ or $$I = JA\cos \theta $$
where, J = current density.
$$ \Rightarrow 5 = J\left( {{4 \over {100}}} \right) \times \cos (60^\circ )$$ [$$\because$$ Given, $$\theta$$ = 60$$^\circ$$]
$$J = 500 \times {1 \over 2}$$ [$$\because$$ cos60$$^\circ$$ = $${1 \over 2}$$]
$$\Rightarrow$$ J = 250 Am$$-$$2
The relation between electric field, current density and resistivity can be given as,
E = $$\rho$$ . J
= 44 $$\times$$ 10$$-$$8 $$\times$$ 250 [$$\because$$ Resistivity, $$\rho$$ = 44 $$\times$$ 10$$-$$8 $$\Omega$$-m]
= 11 $$\times$$ 10$$-$$5 V/m
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