If a copper rod carries a direct current,the | vedclass

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

Question

(C) According to Ampere's Circuital Law,for a cylindrical conductor carrying a steady current $I$ uniformly distributed across its cross-section,the m

Vedclass · Accepted Answer

Both inside and outside the rod

Vedclass · Answer

(C) According to Ampere's Circuital Law,for a cylindrical conductor carrying a steady current $I$ uniformly distributed across its cross-section,the magnetic field $B$ exists at all points.Inside the rod (at a distance $r Outside the rod (at a distance $r > R$),the magnetic field is given by $B = \frac{\mu_0 I}{2 \pi r}$.Therefore,the magnetic field associated with the current exists both inside and outside the rod.

If a copper rod carries a direct current,the magnetic field associated with the current will be

Similar Questions

If $z$ and $\omega$ are two non-zero complex numbers such that $|z\omega| = 1$ and $\text{arg}(z) - \text{arg}(\omega) = \frac{\pi}{2}$,then $\bar{z}\omega$ is equal to

The number of non-trivial solutions of the system $x-y+z=0, x+2y-z=0, 2x+y+3z=0$ is

If $a, b, c$ and $d$ are real numbers such that $a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right]$,then $A^{-1}$ is equal to

What is vegetative propagation? Give two suitable examples.

For one mole of a van der Waals gas,when $b = 0$ and $T = 300 \ K$,the plot of $PV$ versus $1/V$ is shown below. The value of the van der Waals constant $a \ (atm \ L^2 \ mol^{-2})$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module