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Firlakuza [10]
3 years ago
11

A point charge is moving with speed 2 × 107 m/s parallel to the x axis along strait line at y=2 m. At t = 0, the charge is at x

= 0 m. The magnitude of the magnetic field at x = 4 m is B0 at the origin. The magnitude of the magnetic field at at origin when t = 0.15 μs is?
Physics
1 answer:
Oduvanchick [21]3 years ago
4 0

Answer:

B_2  = 4B_1

Explanation:

initial distance of point from the electron is r_1 = 4 m

distance moved by electron is x

x = vt

x = 2*10^{7} *0.1*10^{-6}

x = 2 m

new distance of point from electron is r_2 = 2m

field due to moving charge is

B = \frac{ \mu_o 4vr}{4 \pi r^3}

CHARGE IS INVERSELY PROPOTIONAL TO DISTANCE FROM ELECTRON.

therefore\frac{B_1}{B_2} = [\frac{r_2}{r_1}]^2

                         = [\frac{2}{4}]^2

B_2  = 4B_1

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4 years ago
A crate of books is to be put on a truck with the help of some planks sloping up at 36◦ . The mass of the crate is 80 kg, and th
Dafna11 [192]

Answer:

F= 2.86KN

Explanation:

The attached diagram shows the motion of crate on a single line The forces acting on it have been resolved as shown in the attached figure.The kinetic friction opposes the motion.Now crate is moving with constant speed so the acceleration is constant.

Applying Newton's second law to determine the force F and also substituting the values for F_{k} to eliminate normal force

$\sum_F = ma

Now a_{x}  and  a_{y} components are zero

Horizontal and vertical forces are Force on crate, weight of crate and kinetic friction

$\sum F_{x} = Fcos\theta-f_{k}-mgsin\theta=0

$\sum F_{y} = F_{n}- Fsin\theta-mgcos\theta=0

therefore f_{k} = \mu_{k} F_{n}

Fcos\theta- \mu_{k} F_{n}-mgsin\theta=0  >>>  (1)

F_{n}=Fsin\theta+mgcos\theta

By putting the above equation in equation 1 we get F

F=\frac{mg(sin\theta+\mu_{k}cos\theta) }{cos\theta-\mu_{k}sin\theta }

By putting values we get

F = \frac{(80kg)(9.81\frac{m}{s^{2} })(sin36+(0.8)cos36) }{cos36-(0.8)sin36}

F= 2860.8 N

F= 2.86KN

4 0
4 years ago
What is the normal direction of heat transfer​
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7 0
3 years ago
Read 2 more answers
The magnetic dipole moment of Earth has magnitude 8.00 1022 J/T.Assume that this is produced by charges flowing in Earth’s molte
ziro4ka [17]

Answer:

2.08\cdot 10^9 A

Explanation:

The magnetic dipole moment of a circular coil with a current is given by

\mu = IA

where

I is the current in the coil

A=\pi r^2 is the area enclosed by the coil, where

r is the radius of the coil

So the magnetic dipole moment can be rewritten as

\mu = I\pi r^2 (1)

Here we can assume that the magnetic dipole moment of Earth is produced by charges flowing in Earth’s molten outer core, so by a current flowing in a circular path of radius

r=3500 km = 3.5\cdot 10^6 m

Here we also know that the Earth's magnetic dipole moment is

\mu = 8.0\cdot 10^{22} J/T

Therefore, we can re-arrange eq (1) to find the current that the charges produced:

I=\frac{\mu}{\pi r^2}=\frac{8.00\cdot 10^{22}}{\pi (3.5\cdot 10^6)^2}=2.08\cdot 10^9 A

3 0
4 years ago
A car starts from rest and accelerates at a constant rate. After the car has gone 50 m it has a speed of 21 m/s. What is the acc
juin [17]

Answer:4.41ms-2

Explanation:

From rest U=0,therefore when given S=50m and V=21m/s then you apply the formula of linear motion which is V^2=U^2+2as

Then substitute for the given values and reach the answer e.g 21^2=(0)^2+2×a×50

441m/s=100ma

a=4.41ms-2

6 0
4 years ago
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