An infinitely long conductor PQR is bent to f | vedclass

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

Question

(C) The magnetic field at any point lying on the axis of a current-carrying straight conductor is zero.Case $1$: The current $I$ flows through $PQ$ an

Vedclass · Accepted Answer

$0.67$

Vedclass · Answer

(C) The magnetic field at any point lying on the axis of a current-carrying straight conductor is zero.Case $1$: The current $I$ flows through $PQ$ and $QR$. Point $M$ lies on the extension of $QR$. Therefore,the magnetic field at $M$ due to $QR$ is zero. The magnetic field $H_1$ at $M$ is only due to the segment $PQ$. Let $d$ be the perpendicular distance from $M$ to $PQ$. The magnetic field due to a semi-infinite wire is $H_1 = \frac{\mu_0 I}{4 \pi d}$.Case $2$: The current in $PQ$ is $I$,in $QR$ is $I/2$,and in $QS$ is $I/2$. Point $M$ lies on the extension of $QR$,so the magnetic field due to $QR$ is still zero. The magnetic field at $M$ is now $H_2 = H_{PQ} + H_{QS}$.Since the current in $PQ$ is unchanged,$H_{PQ} = H_1 = \frac{\mu_0 I}{4 \pi d}$.The conductor $QS$ is perpendicular to $PQ$ and $M$ lies on the perpendicular bisector of the line passing through $QS$ if we consider the geometry. However,$M$ is at a distance $d$ from $Q$ along the line $QR$. The magnetic field at $M$ due to a semi-infinite wire $QS$ carrying current $I/2$ is $H_{QS} = \frac{\mu_0 (I/2)}{4 \pi d} = \frac{1}{2} H_1$.Thus,$H_2 = H_1 + \frac{1}{2} H_1 = \frac{3}{2} H_1$.The ratio $H_1/H_2 = H_1 / (\frac{3}{2} H_1) = 2/3 \approx 0.67$.

Similar Questions

Which of the following figures correctly depicts the direction of the magnetic field of a current-carrying coil?

The magnetic field normal to the plane of a coil of $N$ turns and radius $r$ which carries a current $i$ is measured on the axis of the coil at a distance $h$ from the centre of the coil. This is smaller than the field at the centre by the fraction,

The magnetic field at the center of a circular coil carrying current '$I$' for a single turn of a given length of wire is '$B$'. The same wire is bent into a circular coil having two turns. When the same current '$I$' passes through it, the value of the magnetic field becomes:

Assertion : The magnetic field at the centre of the circular coil in the following figure due to the currents $I_1$ and $I_2$ is zero.
Reason : $I_1 = I_2$ implies that the fields due to the current $I_1$ and $I_2$ will be balanced.

The magnitude of the magnetic field at $O$ due to a current-carrying loop as shown in the figure is, where $O$ is the center of two circular portions with radii $1 \, cm$ and $2 \, cm$ respectively. (Take the value of current $I = \frac{1.2}{\pi} \, A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module