New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Magnetics question

2022 · 24 Jun · Shift 1 · Q60

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Physics
Magnetics
2022 · 24 Jun · Shift 1 · Q60

Magnetics question

2022 · 24 Jun · Shift 1 · Q60

JEE MainPhysicsMagneticsMCQ+4 / −1

The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance

{r \over 2}

from the centre is :

A
B/2
B
2B
C
${\left( {{2 \over {\sqrt 5 }}} \right)^3}B$
D
${\left( {{2 \over {\sqrt 3}}} \right)^3}B$

View written solutionFree

Correct answer: C

Magnetic field at the centre of a circular coil

For a circular coil of radius $r$ carrying current $I$ , the magnetic field at the centre is

B_{\text{centre}}=\frac{\mu_0 I}{2r}

Given this is equal to $B$ , so

B=\frac{\mu_0 I}{2r}

Magnetic field on the axis of a circular coil

At a point on the axis at distance $x$ from the centre, the magnetic field is

B_x=\frac{\mu_0 I r^2}{2(r^2+x^2)^{3/2}}

Here, $x=\frac{r}{2}$ . Substituting:

B_x=\frac{\mu_0 I r^2}{2\left(r^2+\left(\frac{r}{2}\right)^2\right)^{3/2}}

Simplify the denominator

r^2+\left(\frac{r}{2}\right)^2=r^2+\frac{r^2}{4}=\frac{5r^2}{4}

So,

\left(\frac{5r^2}{4}\right)^{3/2}=\left(\frac{5}{4}\right)^{3/2}r^3

Thus,

B_x=\frac{\mu_0 I r^2}{2\left(\frac{5}{4}\right)^{3/2}r^3}

B_x=\frac{\mu_0 I}{2r}\cdot \frac{1}{\left(\frac{5}{4}\right)^{3/2}}

But $\frac{\mu_0 I}{2r}=B$ , hence

B_x=B\cdot \left(\frac{4}{5}\right)^{3/2}

Rewrite in the given form

\left(\frac{4}{5}\right)^{3/2}=\left(\frac{2}{\sqrt{5}}\right)^3

Therefore,

B_x=\left(\frac{2}{\sqrt{5}}\right)^3 B

Option check

A: $\dfrac{B}{2}$ — incorrect
B: $2B$ — incorrect
C: $\left(\dfrac{2}{\sqrt{5}}\right)^3 B$ — correct
D: $\left(\dfrac{2}{\sqrt{3}}\right)^3 B$ — incorrect

So the correct answer is Option C.

More from Magnetics

A proton, a deutron and an $α$ -particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is :2022 · MCQ
Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is $6 : 5$ and their respective masses ratio is $9 : 4$ …2022 · MCQ
A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r…2022 · MCQ
A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to2022 · MCQ
A charge particle is moving in a uniform magnetic field $(2 \hat{i} + 3 \hat{j}) T$ . If it has an acceleration of $(α \hat{i} - 4 \hat{j}) m / s^{2}$ , then the value of $α$ will be :2022 · MCQ
$B_{X}$ and $B_{Y}$ are the magnetic fields at the centre of two coils $X$ and $Y$ respectively each carrying equal current. If coil $X$ has 200 turns and $Y$ radius and coil $Y$ …2022 · MCQ
Two concentric circular loops of radii $r_{1} = 30 cm$ and $r_{2} = 50 cm$ are placed in $X - Y$ plane as shown in the figure. A current $I = 7 A$ is flowing through them in the direction as shown in… Includes diagram2022 · MCQ
A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is :2022 · MCQ