Find the magnetic field at point P due to a s | vedclass

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

Question

(D) The magnetic field $B$ at a distance $d$ from a finite straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{4\pi d}(\sin 	heta_1 +

Vedclass · Accepted Answer

$1.5 	imes 10^{-5}\, T$

Vedclass · Answer

(D) The magnetic field $B$ at a distance $d$ from a finite straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{4\pi d}(\sin 	heta_1 + \sin 	heta_2)$.Given,length of wire $AB = 6\, cm$,so the distance from the center to each end is $3\, cm$.The perpendicular distance $d$ from point $P$ to the wire is $4\, cm$ (from the $3-4-5$ triangle geometry).Here,$	heta_1 = 	heta_2 = 	heta$,where $\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{3}{5} = 0.6$.Thus,$B = \frac{\mu_0 I}{4\pi d}(2 \sin 	heta)$.Substituting the values: $B = \frac{(10^{-7} 	imes 4\pi) 	imes 5}{4\pi 	imes (4 	imes 10^{-2})} 	imes 2 	imes \frac{3}{5}$.$B = \frac{10^{-7} 	imes 5}{4 	imes 10^{-2}} 	imes \frac{6}{5} = \frac{10^{-5} 	imes 5}{4} 	imes 1.2 = 1.25 	imes 1.2 	imes 10^{-5} = 1.5 	imes 10^{-5}\, T$.

Find the magnetic field at point $P$ due to a straight line segment $AB$ of length $6\, cm$ carrying a current of $5\, A$. (See figure) $(\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A)$

Similar Questions

State the Biot-Savart law.

The ratio of the magnetic field at the center of a current-carrying circular coil of radius $R$ to the magnetic field at a distance of $3R$ from the center on its axis is:

Write the formula for the magnetic field at a point on the axis of a circular current-carrying loop of radius $R$ at a distance $x$ from its center,where $x >> R$.

The current through $ABC$ and $A'B'C'$ is $I$. What is the magnetic field at $P$? Given $BP = PB' = r$ (Here $C', B', P, B, C$ are collinear).

Vedclass Test Series

Exam Paper Generator

Online Exam Module