Six very long insulated copper wires are boun | vedclass

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

Question

(A) The net current $I_{	ext{net}}$ flowing through the cable is the algebraic sum of the currents in the individual wires:$I_{	ext{net}} = I_1 + I_

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The net current $I_{	ext{net}}$ flowing through the cable is the algebraic sum of the currents in the individual wires:$I_{	ext{net}} = I_1 + I_2 + I_3 + I_4 + I_5 + I_6$$I_{	ext{net}} = 10 - 13 + 10 + 7 - 12 + 18 = 20 	ext{ A}$The magnetic field $B$ at a perpendicular distance $r$ from a long straight wire is given by the formula:$B = \frac{\mu_0 I_{	ext{net}}}{2\pi r}$Given $r = 10 	ext{ cm} = 0.1 	ext{ m}$ and $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$:$B = \frac{4\pi 	imes 10^{-7} 	imes 20}{2\pi 	imes 0.1}$$B = \frac{2 	imes 10^{-7} 	imes 20}{0.1}$$B = 400 	imes 10^{-7} 	ext{ T} = 4 	imes 10^{-5} 	ext{ T}$$B = 40 \mu	ext{T}$

Similar Questions

The magnetic induction in air at a point $1\,cm$ away from a long straight wire that carries a current of $1\,A$ will be:

$20 \, A$ current is flowing in a long straight wire. The intensity of the magnetic field at a distance of $20 \, cm$ from the wire will be:

The magnetic field at the centre of a circular coil of radius $R$,carrying current $2 \ A$ is $B_1$. The magnetic field at the centre of another coil of radius $3R$ carrying current $4 \ A$ is $B_2$. The ratio $B_1: B_2$ is

Magnetic field at a distance $r$ from an infinitely long straight conductor carrying a steady current varies as

Vedclass Test Series

Exam Paper Generator

Online Exam Module