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adelina 88 [10]

3 years ago

13

A vertical wire carries a current straight down. To the east of this wire, the magnetic field points: A) toward the east. B) tow

ard the west. C) toward the north. D) downward. E) toward the south.

2 answers:

geniusboy [140]3 years ago

6 0

Answer:

option E

Explanation:

the correct answer is option E

the direction of magnetic field will be found out with the help of right hand rule.

Put you palm in the direction of electric field and curl your finger in the direction of magnetic field which east direction.

now, the direction shown by the thumb will be the direction of magnetic field which comes out to be toward South direction.

Marizza181 [45]3 years ago

4 0

Answer:

answer is E) towards the south

Explanation:

the direction of magnetic field is found by right hand rule, if you curl your fingers to the east the thumb with point towards south.

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An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect

klio [65]

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = \frac{1}{2}mv^2$

Making v the subject

$v = \sqrt{[\frac{2 \Delta V * q }{m}] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $\frac{mv^2}{r}$ ] and this is mathematically represented as

$Bqv = \frac{mv^2}{r}$

making B the subject

$B = \frac{mv}{rq}$

r is the radius with a value = 5.4cm = $= \frac{5.4}{100} = 5.4*10^{-2} m$

Substituting values

$B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}$

$= 0.0048 T$

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4 years ago