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3 years ago

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Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45

A. Determine the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current.

1 answer:

serious [3.7K]3 years ago

5 0

Explanation:

It is given that,

The separation between two parallel wires, r = 11 cm = 0.11 m

Current in wire 1, $q_1=54\ A$

Current in wire 2, $q_2=45\ A$

Length of wires, l = 4.3 m

We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :

$\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N$

So, the magnetic force on a 4.3 m length of the wire on both of currents is F=0.0189 N.

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A) 21.2 kg.m/s at 39.5 degrees from the x-axis

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Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

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3 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

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A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

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3 years ago

a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres

Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

$\text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}$

Similarly,

$\text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}$

We know the below formulas,

$\text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}$

$\text {volume of wire}=\pi \times r^{2} \times l$

When equating both the equations, we can find length of wire as below, where $\pi=\frac{22}{7}$

$\frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l$

$\frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l$

The $\pi$ value gets cancelled as common on both sides, we get

$\frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l$

The $10^{-6}$ value gets cancelled as common on both sides, we get

$l=4 \times 9=36 m$

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3 years ago