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deff fn [24]
4 years ago
9

A charged particle moving along the +x-axis enters a uniform magnetic field pointing along the +z-axis. A uniform electric field

is also present. Due to the combined effect of both fields, the particle does not change its velocity. What is the direction of the electric field?
Physics
1 answer:
Brums [2.3K]4 years ago
3 0

Answer:

the electric field direction should be in positive y axis

Explanation:

Lets assume that charge on particle is positive and it isequal to +q

First calculate the magnetic force on it

F_B = q V\times B =qVBsin\theta

for direction

use Right Hand Rule which will give the direction and by using his the direction will come towards negative y axis.

As given in the question that charge particle does not change their velocity so we need to apply electric field in such a way that electric force direction should be opposite to the magnectic field.

and magnitude should be same as magnectic force and also direcion of electric force depend on the direction of elecric field when charge is positive because electric force F_E = qE

Hence the electric field direction should be in positive y axis

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Define s.i unit of force explanation
kvv77 [185]
<h3><u>Answer</u> :</h3>

◈ As per newton's second law of motion, Force is defined as the product of mass and acceleration.

Mathematically,

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Unit of mass : kg

Unit of acceleration : m/s²

Therefore,

Unit of force ➠ <u>kg m/s²</u>

SI unit : <u>N (newton)</u> or <u>kg m/s²</u>

3 0
4 years ago
A radar station detects an airplane approaching directly from the east. at first observation, the range to the plane is d1 = 354
PolarNik [594]
The airplane is tracked for another 123° in the vertical eastwest plane
8 0
3 years ago
Can someone please explain how to solve 13?
frozen [14]

Hello!

Everything you've done so far seems to be correct. However, you need to get rid of the sin. To do this, use ssin^{-1}.

This is known as the inverse of sin, which is where you go from a fraction to the actual angle itself.

When you run sin^{-1}(1/5) through a calculator, you get about 11.537 degrees, or θ = 11.537.

To verify this, all you need to do is run sin (11.537), and the calculator returns 0.2, which is what we're looking for.

Hope this helps!

8 0
3 years ago
A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is 5) its radius? (The value of o is 4 ×
Elan Coil [88]
The self-inductance of a solenoid is given by:
L= \frac{\mu_0 N^2 A}{l}
where
\mu_0 is the vacuum permeability
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid

For the solenoid in our problem, N=3000, l=70.0 cm=0.70 m and the self-inductance is L=25.0 mH=0.025 H, therefore the cross-sectional area is
A= \frac{Ll}{\mu N^2}= \frac{(0.025 H)(0.70 m)}{(4\pi \cdot 10^{-7}N/A^2)(3000)^2}= 1.55 \cdot 10^{-3}m^2
And since the area is related to the radius by
A=\pi r^2
The radius of the solenoid is
r= \sqrt{ \frac{A}{\pi} } = \sqrt{ \frac{1.55 \cdot 10^{-3} m^2}{\pi} } =0.022 m=2.2 cm

6 0
3 years ago
A circular 10-turn coil with a radius of 5.0 cm carries a current of 5 A. It lies in the xy plane in a uniform magnetic field =
Tju [1.3M]

Answer:

U = – 0.12J

Explanation:

Given N = 10 turns, I = 5A, r = 5×10-²m

B^ = 0.05 T iˆ+ 0.3 T kˆ

Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T

Area = πr² = π(5×10-²)² = 7.85×10-³m²

Magnetic moment μ = NIA

μ = 10×5×7.85×10-³ = 0.3925Am²

U = -μ•B = –0.3925×0.304 = –0.12J

The sign is negative because the magnetic moment is aligned with the magnetic field.

3 0
3 years ago
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