Two infinitely long wires each carrying curre | vedclass

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

Question

(D) In the given arrangement,the net magnetic field at point $P$ is the vector sum of the magnetic fields produced by the straight wire segments $AB$,

Vedclass · Accepted Answer

$\frac{\mu_0 I}{2 \pi r}$

Vedclass · Answer

(D) In the given arrangement,the net magnetic field at point $P$ is the vector sum of the magnetic fields produced by the straight wire segments $AB$,$CD$,$EF$,$GH$ and the circular arc segments $BC$ and $FG$.$1$. The point $P$ lies on the axis of the straight wires $EF$ and $GH$. Therefore,the magnetic field due to these segments at $P$ is zero: $B_{EF} = B_{GH} = 0$.$2$. The magnetic field due to a semi-infinite wire at a distance $r$ from its end is $B = \frac{\mu_0 I}{4 \pi r}$.$3$. The magnetic field due to the straight segment $AB$ at $P$ is $B_{AB} = \frac{\mu_0 I}{4 \pi r}$ (directed into the page).$4$. The magnetic field due to the straight segment $CD$ at $P$ is $B_{CD} = \frac{\mu_0 I}{4 \pi r}$ (directed into the page).$5$. The magnetic field due to the quarter-circular arc $BC$ at $P$ is $B_{BC} = \frac{\mu_0 I}{8 r}$ (directed out of the page).$6$. The magnetic field due to the quarter-circular arc $FG$ at $P$ is $B_{FG} = \frac{\mu_0 I}{8 r}$ (directed into the page).$7$. The net magnetic field is $B_{net} = B_{AB} + B_{CD} + B_{FG} - B_{BC} = \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{8 r} - \frac{\mu_0 I}{8 r} = \frac{2 \mu_0 I}{4 \pi r} = \frac{\mu_0 I}{2 \pi r}$.Thus,the correct option is $(d)$.

Column $I$	Column $II$
$(A)$ Two parallel wires with current in the same direction,$P$ is the midpoint.	$(p)$ The magnetic fields $(B)$ at $P$ due to the currents in the wires are in the same direction.
$(B)$ Two coaxial circular loops with current in the same direction,$P$ is the midpoint on the axis.	$(q)$ The magnetic fields $(B)$ at $P$ due to the currents in the wires are in opposite directions.
$(C)$ Two coplanar circular loops with current in opposite directions,$P$ is the midpoint.	$(r)$ There is no magnetic field at $P$.
$(D)$ Two concentric coplanar circular loops with current in the same direction,$P$ is the common center.	$(s)$ The wires repel each other.

Two infinitely long wires each carrying current $I$ along the same direction are made into the geometry as shown in the figure below. The magnetic field at the point $P$ is

Similar Questions

The dimensional formula of $\frac{1}{2} \mu_0 H^2$ (where $\mu_0$ is the permeability of free space and $H$ is the magnetic field intensity) is:

$A$ current-carrying closed loop in the form of a right-angled isosceles triangle $ABC$ is placed in a uniform magnetic field acting along $AB.$ If the magnetic force on the arm $BC$ is $\vec F,$ the force on the arm $AC$ is

$A$ closed loop $PQRS$ carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments $PS, SR$ and $RQ$ are $F_1, F_2$ and $F_3$ respectively and are in the plane of the paper and along the directions shown,the force on the segment $QP$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module