A long straight metal rod of radius b has a v | vedclass

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

Question

(A) The magnetic field at a point inside a solid current-carrying cylinder with uniform current density $J$ is given by $B = \frac{{\mu _0 J r}}{2}$,w

Vedclass · Accepted Answer

$\frac{{\mu _0 ic}}{{2\pi (b^2 - a^2)}}$

Vedclass · Answer

(A) The magnetic field at a point inside a solid current-carrying cylinder with uniform current density $J$ is given by $B = \frac{{\mu _0 J r}}{2}$,where $r$ is the distance from the axis.To solve this,we use the principle of superposition. We consider the rod with the hole as a solid cylinder of radius $b$ carrying current density $J$ in one direction,and a smaller cylinder of radius $a$ carrying current density $J$ in the opposite direction.The total current $i$ is given by $i = J \cdot \pi (b^2 - a^2)$,so $J = \frac{i}{{\pi (b^2 - a^2)}}$.At the axis of the hole (point $C$),the magnetic field due to the large cylinder is $B_1 = \frac{{\mu _0 J c}}{2}$ (directed perpendicular to $OC$).The magnetic field due to the hole (the smaller cylinder) at its own axis is $B_2 = 0$.The net magnetic field at $C$ is $B = B_1 - B_2 = \frac{{\mu _0 J c}}{2}$.Substituting $J$,we get $B = \frac{{\mu _0 i c}}{{2\pi (b^2 - a^2)}}$.

$A$ long straight metal rod of radius $b$ has a very long hole of radius $a$ drilled parallel to the rod axis as shown in the figure. If the rod carries a total current $i$,find the value of magnetic induction on the axis of the hole,where $OC = c$.

Similar Questions

Two long parallel wires $P$ and $Q$ are held perpendicular to the plane of the paper with a distance of $5 \; m$ between them. If $P$ and $Q$ carry currents of $2.5 \; A$ and $5 \; A$ respectively in the same direction,then the magnetic field at a point half-way between the wires is:

If a battery of $12 \text{ V}$ is connected across the diametrically opposite points $A$ and $B$ of a conducting ring of radius $R$ and the current drawn from the battery is $I$,then the magnetic field produced at the centre of the ring due to the ring is . . . . . . .

The magnetic flux passing through the rectangular loop is proportional to:

$A$ straight wire carrying current $I$ is bent into a semi-circular arc of radius $r$,as shown. The magnitude of the magnetic field at point $O$ due to the semi-circular arc is ($\mu_{0} =$ permeability of free space).

Vedclass Test Series

Exam Paper Generator

Online Exam Module