Two long straight wires are placed along the | vedclass

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

Question

(C) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.For a point $(x, y)$ in the first q

Vedclass · Accepted Answer

$y=\left(\frac{I_1}{I_2}\right) x$

Vedclass · Answer

(C) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.For a point $(x, y)$ in the first quadrant,the magnetic field due to the wire along the $x$-axis (carrying current $I_1$) is $B_1 = \frac{\mu_0 I_1}{2 \pi y}$ (directed out of the plane).The magnetic field due to the wire along the $y$-axis (carrying current $I_2$) is $B_2 = \frac{\mu_0 I_2}{2 \pi x}$ (directed into the plane).For the net magnetic field to be zero,the magnitudes must be equal: $B_1 = B_2$.$\frac{\mu_0 I_1}{2 \pi y} = \frac{\mu_0 I_2}{2 \pi x}$.Simplifying this,we get $\frac{I_1}{y} = \frac{I_2}{x}$,which implies $y = \left(\frac{I_1}{I_2}\right) x$.

Two long straight wires are placed along the $x$-axis and $y$-axis. They carry currents $I_1$ and $I_2$ respectively. The equation of the locus of points with zero magnetic induction in the magnetic field produced by them is:

Similar Questions

$A$ vertical wire kept in the $Z-X$ plane carries a current from $Q$ to $P$ (see figure). The magnetic field due to the current will have the direction at the origin $O$ along

$A$ current of $0.1\, A$ circulates around a coil of $100$ turns and having a radius equal to $5\, cm$. The magnetic field set up at the centre of the coil is $({\mu _0} = 4\pi \times {10^{ - 7}}\,T\cdot m/A)$.

Which of the following graphs represents the variation of magnetic field $B$ with perpendicular distance $r$ from an infinitely long,straight conductor carrying current?

Which of the following graphs correctly represents the variation of magnetic field $B$ with distance $R$ from a long straight current-carrying conductor?

Vedclass Test Series

Exam Paper Generator

Online Exam Module