All categories

4 years ago

In which direction is the magnetic force acting on the charge?

Physics

1 answer:

adell [148]4 years ago

6 0

Force on a moving charge inside magnetic field is given by

$F = q(v x B)$

So as per right hand rule if we put our fingers of right hand along the velocity of charge and then curl it towards the magnetic field direction then thumb will show the direction of force on moving charge.

Above rule is known as right hand rule.

If we use the same for the given condition then velocity is given out of the plane and magnetic field is downwards then as per right hand rule direction of force on the moving charge will be "Towards Right"

You might be interested in

A shopping cart is given an initial velocity of 2 m/s and undergoes a constant acceleration of 4 m/s2. What is the magnitude of

Rasek [7]

Answer:

84 meters

Explanation:

Using these two kinematic equations : Vf= Vi +at and

D= [(Vi+Vf)/2]*t

Plugging in what we know into the first equation to solve for the Final velocity, you'll get Vf= 2 m/s + 4(m/s^2)*6s = 26m/s

plug time vi and vf into the second equation to solve for displacement.

6 0

3 years ago

All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will

VikaD [51]

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

$v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}$

since $v_y = 0$ at the top of its trajectory. Plugging in the numbers,

$t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}$

8 0

3 years ago

What is the mass of .5 newtons(N)?

ddd [48]

Since the mass is 5 grams the acceleration is: 4000 m/s^2

But if the mass is 5 Kilograms the acceleration is: 4 m/s^2

7 0

3 years ago

At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particu

Ne4ueva [31]

Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = $E_{h}$

Let the vertical electric field component = $E_{v}$

The formula for light intensity is given by:

$I = \frac{E_{m} ^{2} }{2c \mu}$ ..............................(1)

$E_{m}$ is the resolution of the vertical and horizontal components, $E_{h} and E_{v}$

$E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}$ ..................(2)

Light intensity before the glasses were put on:

$I_{1} = \frac{E_{m} ^{2} }{2c \mu_{1} }$ .............................(3)

Put equation (2) into equation (3)

$I_{1} = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }$ .............................(4)

After the glasses were put on the horizontal component vanishes, i.e. $E_{h} = 0$

$I_{2} = \frac{ E_{v} ^{2}}{2c \mu_{2} }$ ...................................(5)

Divide equation (5) by equation (4)

$\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$ ...............................(6)

But $E_{h} = 2.3E_{v}$ ......................(7)

Insert equation (7) into (6)

$\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{(2.3E_{v})^{2} + E_{v} ^{2} } \\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{5.29E_{v}^{2} + E_{v} ^{2} }\\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\$