The fractional change in the magnetic field i | vedclass

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

Question

(D) The magnetic field on the axis of a current-carrying coil of radius $a$ at a distance $r$ from the centre is given by $B_{	ext{axis}} = \frac{\mu

Vedclass · Accepted Answer

$\frac{3}{2} \frac{r^{2}}{a^{2}}$

Vedclass · Answer

(D) The magnetic field on the axis of a current-carrying coil of radius $a$ at a distance $r$ from the centre is given by $B_{	ext{axis}} = \frac{\mu_{0} i a^{2}}{2(a^{2} + r^{2})^{3/2}}$.The magnetic field at the centre of the coil is $B_{	ext{centre}} = \frac{\mu_{0} i}{2a}$.The fractional change in magnetic field is defined as $\frac{B_{	ext{centre}} - B_{	ext{axis}}}{B_{	ext{centre}}} = 1 - \frac{B_{	ext{axis}}}{B_{	ext{centre}}}$.Substituting the expressions: $1 - \frac{\frac{\mu_{0} i a^{2}}{2(a^{2} + r^{2})^{3/2}}}{\frac{\mu_{0} i}{2a}} = 1 - \frac{a^{3}}{(a^{2} + r^{2})^{3/2}} = 1 - \left(1 + \frac{r^{2}}{a^{2}}ight)^{-3/2}$.Using the binomial approximation $(1 + x)^{n} \approx 1 + nx$ for $x $1 - (1 - \frac{3}{2} \frac{r^{2}}{a^{2}}) = \frac{3}{2} \frac{r^{2}}{a^{2}}$.

The fractional change in the magnetic field intensity at a distance $r$ from the centre on the axis of a current-carrying coil of radius $a$ to the magnetic field intensity at the centre of the same coil is: (Take $r << a$)

Similar Questions

In the current-carrying conductor $(AOCDEFG)$ as shown,the magnetic induction at the point $O$ is ($R_1$ and $R_2$ are radii of arcs $CD$ and $EF$ respectively,$I$ = current in the loop,$\mu_0$ = permeability of free space).

Shown in the figure is a conductor carrying a current $I$. The magnetic field intensity at the point $O$ (common centre of all the three arcs) is

$A$ current $I$ is flowing in a conductor $PQRST$ as shown in the figure. The radius of the curved path $QRS$ is $r$ and the length of the straight portions $PQ$ and $ST$ is very large. The magnetic field at the centre $O$ of the curved part is $(\mu_0 = \text{permeability of free space})$

$A$ coil having $9$ turns carrying a current produces a magnetic field $B_1$ at the centre. Now,the coil is rewound into $3$ turns carrying the same current. Then,the magnetic field at the centre $B_2$ is:

Two wires $A$ and $B$ are of lengths $40 \ cm$ and $30 \ cm$. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r$. $A$ current $i_1$ is passed through $A$ and $i_2$ through $B$. To have the same magnetic inductions at the centre, the ratio of $i_1: i_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module