Two identical long conducting wires AOB and C | vedclass

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

Question

(C) The magnetic field due to a long straight wire at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{2\pi d}$.For wire $AOB$ carrying cu

Vedclass · Accepted Answer

$\frac{\mu_0}{2\pi d} (I_1^2 + I_2^2)^{1/2}$

Vedclass · Answer

(C) The magnetic field due to a long straight wire at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{2\pi d}$.For wire $AOB$ carrying current $I_1$,the magnetic field at point $P$ (at distance $d$ along the perpendicular axis) is $B_1 = \frac{\mu_0 I_1}{2\pi d}$.For wire $COD$ carrying current $I_2$,the magnetic field at point $P$ is $B_2 = \frac{\mu_0 I_2}{2\pi d}$.Since the wires are perpendicular to each other,the magnetic field vectors $\vec{B}_1$ and $\vec{B}_2$ at point $P$ are also perpendicular to each other.The resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.Substituting the values: $B = \sqrt{\left( \frac{\mu_0 I_1}{2\pi d} \right)^2 + \left( \frac{\mu_0 I_2}{2\pi d} \right)^2}$.$B = \frac{\mu_0}{2\pi d} \sqrt{I_1^2 + I_2^2} = \frac{\mu_0}{2\pi d} (I_1^2 + I_2^2)^{1/2}$.

Similar Questions

$A$ long straight wire of circular cross-section of radius $a$ is carrying a steady current. The current is distributed uniformly across the cross-section of the wire. The ratio of the magnetic fields at points $0.5a$ and $1.5a$ from the centre of the wire is

$A$ straight wire of finite length carrying current $I$ subtends an angle of $60^{\circ}$ at point $P$ as shown. The magnetic field at $P$ is

If a positive ion is moving away from an observer with some acceleration,then the lines of force of magnetic induction will be

The magnetic field at the point of intersection of diagonals of a square wire loop of side $L$ carrying a current $I$ is

$A$ circular loop and an infinitely long straight conductor carry equal currents,as shown in the figure. The net magnetic field at the centre of the loop is $B_1$,when the current in the loop is clockwise and $B_2$ when the current in the loop is anti-clockwise. Then $\frac{B_1}{B_2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module