Physics, 15.04.2020 23:54 zimbabwayjerry

5. An infinite wire is 3.00 cm away from and parallel to a 10 cm long wire, and each has a current of 2.50 A in the same direction (for example the top wire of the rectangle in Fig. 4). (a) What is the magnetic field from the infinite wire at the position of the other wire

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Answer from: mikaydat

B=1.04*10^{-4}T

Explanation:

An infinite wire produces a magnetic field that can be computed with the formula:

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

where m0 is the magnetic permeability of vacuum (4pi*10^{-7}Tm/A), I is the current of the wire (2.50A) and r is the perpendicular distance to the wire in which B is calculated (3cm=3*10^{-2}m). By replacing we obtain:

$B=\frac{(4\pi*10^{-7}Tm/A)(2.50A)}{3*10^{-2}m}=1.04*10^{-4}T$

hope this helps!!

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Answer from: Quest

Answerthe force between the charges will be f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges: $f = \dfrac{k q_1 q_2}{r^2}$ k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then $\begin{aligned} f & = \dfrac{k q_1 q_2}{r^2} \\ f_{3r} & = \dfrac{k q_1 q_2}{(3r)^2} \\ & = \dfrac{k q_1 q_2}{3^2 \cdot r^2} \\ & = \frac{1}{9} \cdot \boxed{\dfrac{k q_1 q_2}{r^2} } \\ & = \frac{1}{9} f & & \left(\text{\footnotesize since $ f = \tfrac{k q_1 q_2}{r^2}$}\right) \end{aligned}$

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5. An infinite wire is 3.00 cm away from and parallel to a 10 cm long wire, and each has a current o...

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