A cylindrical conductor of radius R carries a | vedclass

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

Question

(B) For a cylindrical conductor of radius $R$, the magnetic field inside at a distance $r < R$ is given by $B_{in} = \frac{\mu_0 i r}{2 \pi R^2}$.Give

Vedclass · Accepted Answer

$\frac{8}{3} \ T$

Vedclass · Answer

(B) For a cylindrical conductor of radius $R$, the magnetic field inside at a distance $r Given that the point is $R/4$ inside from the surface, the distance from the center is $r = R - R/4 = 3R/4$.Thus, $B_{in} = \frac{\mu_0 i (3R/4)}{2 \pi R^2} = \frac{3 \mu_0 i}{8 \pi R} = 10 \ T$.From this, we find $\frac{\mu_0 i}{2 \pi R} = \frac{10 	imes 4}{3} = \frac{40}{3} \ T$.For a point outside at a distance $4R$ from the surface, the distance from the center is $r' = R + 4R = 5R$.The magnetic field outside is $B_{out} = \frac{\mu_0 i}{2 \pi r'} = \frac{\mu_0 i}{2 \pi (5R)} = \frac{1}{5} \left( \frac{\mu_0 i}{2 \pi R} ight)$.Substituting the value, $B_{out} = \frac{1}{5} 	imes \frac{40}{3} = \frac{8}{3} \ T$.

$A$ cylindrical conductor of radius $R$ carries a current $i$. The value of the magnetic field at a point, which is $R/4$ distance inside from the surface, is $10 \ T$. Find the value of the magnetic field at a point, which is $4R$ distance outside from the surface.

Similar Questions

$A$ circular arc of radius $r$ carrying current $I$ subtends an angle $\frac{\pi}{8}$ at its centre. The radius of the metal wire is uniform. The magnetic induction at the centre of the circular arc is ($\mu_0 =$ permeability of free space).

In the diagram,$I_{1}$ and $I_{2}$ are the strengths of the currents in the loop and straight conductor,respectively. Given $OA = AB = R$. The net magnetic field at the centre $O$ is zero. Then the ratio of the currents in the loop and the straight conductor is:

$A$ circular coil carrying current has radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{27}$th of its value at the centre of the coil is:

$A$ particle is moving with velocity $\overrightarrow{v} = \hat{i} + 3\hat{j}$ and it produces an electric field at a point given by $\overrightarrow{E} = 2\hat{k}$. It will produce a magnetic field at that point equal to (all quantities are in $SI$ units):

The magnetic induction at the centre $O$ in the figure shown is

Vedclass Test Series

Exam Paper Generator

Online Exam Module