A conductor of length 1.5 m and area of cros | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given: Length $l = 1.5 \ m$,Area $A = 3 	imes 10^{-5} \ m^2$,Resistance $R = 15 \ \Omega$,Electric field $E = 21 \ Vm^{-1}$.We know that current

Vedclass · Accepted Answer

$0.7 	imes 10^5 \ Am^{-2}$

Vedclass · Answer

(D) Given: Length $l = 1.5 \ m$,Area $A = 3 	imes 10^{-5} \ m^2$,Resistance $R = 15 \ \Omega$,Electric field $E = 21 \ Vm^{-1}$.We know that current density $J = \sigma E$,where $\sigma$ is the conductivity.Since $\sigma = \frac{1}{ho} = \frac{l}{RA}$,we can write $J = \frac{l}{RA} 	imes E$.Substituting the values:$J = \frac{1.5 	imes 21}{15 	imes 3 	imes 10^{-5}}$$J = \frac{31.5}{45 	imes 10^{-5}}$$J = 0.7 	imes 10^5 \ Am^{-2}$.

$A$ conductor of length $1.5 \ m$ and area of cross-section $3 \times 10^{-5} \ m^2$ has an electrical resistance of $15 \ \Omega$. The current density in the conductor for an electric field of $21 \ Vm^{-1}$ is

Similar Questions

During the reaction of permanganate with thiosulphate,the change in oxidation state of manganese occurs by a value of $3$. Identify which of the below media will favour the reaction.

Which component of a lichen is heterotrophic?

If $x^y=y^x$, then $x(x-y \log x) \frac{d y}{d x}$ is equal to :

Which of the following is related to genetic engineering?

The focal chord to $y^2 = 16x$ is tangent to $(x - 6)^2 + y^2 = 2$. Then the possible values of the slope of this chord are

Vedclass Test Series

Exam Paper Generator

Online Exam Module