Two identical circular loops P and Q each of | 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 circular loop of radius $r$ at a distance $x$ from its center is given by $B = \frac{\mu_0 I r^2}{2(r^2 + x^2)

Vedclass · Accepted Answer

$\frac{3\mu_{0}I}{4\sqrt{2}r}$ towards $Q$

Vedclass · Answer

(D) The magnetic field on the axis of a circular loop of radius $r$ at a distance $x$ from its center is given by $B = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$.Here, for both loops $P$ and $Q$, the distance from the center $O$ is $x = r$.Thus, the magnetic field due to loop $P$ $(B_P)$ at $O$ is $B_P = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2\sqrt{2}r^3)} = \frac{\mu_0 I}{4\sqrt{2}r}$.Since the current in $P$ is clockwise as seen from $O$, the magnetic field $B_P$ points towards $P$ (away from $Q$).The magnetic field due to loop $Q$ $(B_Q)$ at $O$ is $B_Q = \frac{\mu_0 (4I) r^2}{2(r^2 + r^2)^{3/2}} = \frac{4\mu_0 I r^2}{2(2\sqrt{2}r^3)} = \frac{4\mu_0 I}{4\sqrt{2}r}$.Since the current in $Q$ is also clockwise as seen from $O$, the magnetic field $B_Q$ points towards $Q$ (away from $P$).The net magnetic field $B_{net} = B_Q - B_P = \frac{4\mu_0 I}{4\sqrt{2}r} - \frac{\mu_0 I}{4\sqrt{2}r} = \frac{3\mu_0 I}{4\sqrt{2}r}$ towards $Q$.

Two identical circular loops $P$ and $Q$ each of radius $r$ are lying in parallel planes such that they have a common axis. The currents through $P$ and $Q$ are $I$ and $4I$ respectively in the clockwise direction as seen from $O$. The net magnetic field at $O$ is:

Similar Questions

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

Which rule can be used to determine the direction of the magnetic field due to a current-carrying loop?

The radius of a circular current-carrying coil is $R$. At what distance from the centre of the coil on its axis,the intensity of the magnetic field will be $\frac{1}{2 \sqrt{2}}$ times that at the centre?

Find the magnetic field at point $P$ due to a straight line segment $AB$ of length $6\, cm$ carrying a current of $5\, A$. (See figure) $(\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A)$

Apply the Biot-Savart law to find the magnetic field due to a circular current-carrying loop at a point on the axis of the loop.

Vedclass Test Series

Exam Paper Generator

Online Exam Module