A current of 5A is passing through a non-linear magnesium wire of cross – section 0.04m². At every point the direction of current density is at an angle of 60° ;with unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is: (Resistivity of magnesium $ρ = 44 \times 1 0^{- 8} Ω m)$

Option 1 -

11 * 10^-7 V/m

Option 2 -

11 * 10^-2 V/m

Option 3 -

11 * 10^-5 V/m

Option 4 -

11 * 10^-³ V/m

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

$l = J A c o s θ \Rightarrow 5 = J \times 0.04 \times c o s 60 ° \Rightarrow J = \frac{5}{0.02} = 250 A / m^{2}$

$J = \frac{E}{ρ} \Rightarrow E = ρ J = 44 \times 1 0^{- 8} \times 250 = 11 \times 1 0^{- 5} V / m$

0 Upvotes 0 Downvotes

Similar Questions for you

A wire of length 'l' and resistance $100 Ω$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

alok kumar singh

Divided into 10 parts

$R = \frac{ρ l}{A}$

$R^{'} = \frac{ρ l}{10 A} = \frac{R}{10}$

$R_{S} = 5 \times \frac{R}{10} [series]$

$R_{S} = 50$

$R_{P} = \frac{R}{50} [parallel]$

$R_{eq} = R_{S} + R_{P}$

$= 52 Ω$

A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in V) shown in the figure is:

(Take $\frac{1}{4 π ε_{0}} = 9 \times 1 0^{9}$ SI units)

alok kumar singh

For uniformly charged spherical shell,

$V = \frac{k q}{R} (For r \leq R)$

$∴ V_{C} = V_{P}$

$V_{C} - V_{P} = Zero$

A uniform heating wire of resistance 36 $Ω$ is connected across a potential difference of 240V. The wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is________.

alok kumar singh

In first condition R₁ = 36 $Ω$

In second condition R₂ = 18 $Ω$

$P_{1} = \frac{V^{2}}{R_{1}} = \frac{{(240)}^{2}}{36}$

$P_{2} = \frac{V^{2}}{R_{2}} + \frac{V^{2}}{R_{2}} = \frac{{(240)}^{2}}{18} + \frac{{(240)}^{2}}{18}$

$P_{2} = \frac{{(240)}^{2}}{9}$

So $\frac{P_{1}}{P_{2}} = \frac{{(240)}^{2} / 36}{{(240)}^{2} / 9} = \frac{1}{4}$

x = 4

The terminal voltage of the battery, whose emf is 10 V and internal resistance $1 Ω$ , when connected through an external resistance of $4 Ω$ as shown in the figure is: