Physics, 29.06.2019 12:30 nikkiwoodward1ovgszp

A)find the minimum magnetic field needed to exert a 5.7 fn force on an electron moving at 23mm/s .b)find the field strength required if the field were at 45 degrees to the electron's velocity.

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Answer from: trevorhenyan51

The minimum magnetic field needed to exert the given force is $1.55 \times 10^{-3} \ T$

When the field is 45 degrees to the electron's speed, the magnetic field strength is $2.19\times 10^{-3} \ T$

The given parameters;

force exerted on the electron, F = 5.7 Nspeed of the electron, v = 23 M m/s

The minimum magnetic field is calculated as follows;

$F = qvB \\\\B = \frac{F}{qv} \\\\B = \frac{5.7\times 10^{-15}}{(1.602\times 10^{-19})(23\times 10^6)} \\\\B = 1.55 \times 10^{-3} \ T$

When the field is 45 degrees to the electron's speed, the magnetic field strength is calculated as follows;

$B = \frac{F}{q\times v \times sin(45)} = \frac{5.7 \times 10^{-15}}{1.602 \times 10^{-19} \times 23 \times 10^6 \times sin(45)} = 2.19 \times 10^{-3} \ T$

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Answer from: solomander4

a) $1.55\cdot 10^{-3} T$

The magnetic force exerted on a charged particle in motion is given by:

$F=qvB sin \theta$

where

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

$\theta$ is the angle between the direction of v and B

The expression can be rewritten as

$B=\frac{F}{qv sin \theta}$

We see that the minimum magnetic field needed is the one for which $sin \theta=1$ , so with $\theta=90^{\circ}$ . In this problem, we have:

$q=1.6\cdot 10^{-19} C$ (charge of the electron)

$f=5.7 fN=5.7\cdot 10^{-15}N$ is the force

$v=23 Mm/s = 23\cdot 10^6 m/s$ is the electron's velocity

Substituting, we find

$B=\frac{5.7\cdot 10^{-15}N}{(1.6\cdot 10^{-19} C)(23\cdot 10^6 m/s) sin 90^{\circ}}=1.55\cdot 10^{-3} T$

b) $2.19\cdot 10^{-3} T$

In this case, the field is at 45 degrees to the electron's velocity, so we have

$\theta=45^{\circ}$

Therefore, the field strength required to obtain a force of

$f=5.7 fN=5.7\cdot 10^{-15}N$ is the force

will be equal to

$B=\frac{F}{qv sin \theta}=\frac{5.7\cdot 10^{-15}N}{(1.6\cdot 10^{-19} C)(23\cdot 10^6 m/s) sin 45^{\circ}}=2.19\cdot 10^{-3} T$

Answer from: Quest

the grams of oxygen gas that can be produced is 20.6 grams of o₂

calculation

2liclo₃ → 2 licl + 3 o₂

step 1: find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5 +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281 moles

step 2: use the mole ratio to calculate the moles of o2

from equation above, liclo₃: o₂ is 2: 3

therefore the moles of o₂=0.4281 moles x3/2 =0.6422 moles

step 3: find mass of o₂

mass= moles x molar mass

from periodic table the molar mass of o₂ = 16 x2= 32 g/mol

mass=0.6422 moles x 32 g/mol = 20.6 grams of o₂

Answer from: Quest

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