Ask question

Ask question

All categories

2 years ago

15

A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh

en it is at an angle of 30° with respect to the field?

2 answers:

zheka24 [161]2 years ago

7 0

Answer:

Magnetic force, F = 0.24 N

Explanation:

It is given that,

Current flowing in the wire, I = 4 A

Length of the wire, L = 20 cm = 0.2 m

Magnetic field, B = 0.6 T

Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

$F=ILB\ sin\theta$

$F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)$

F = 0.24 N

So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.

lara31 [8.8K]2 years ago

7 0

Explanation:

The given data is as follows.

length = 20 cm, current = 4 A

B = 0.6 T, $\theta$ = 30°

Hence, formula to calculate the force acting on wire is as follows.

F = $IlBsin \theta$

Now, putting the given values into the above formula as follows.

F = $IlBsin \theta$

= $4 A \times 20 cm \times \frac{10^{-2}}{1 cm} \times 0.6 T \times Sin(30^{o})$

= 0.24 N

= 0.2 N

thus, we can conclude that force on the wire is 0.2 N.

You might be interested in

A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. in the water what angle

alexira [117]

<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>

According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.

Therefore; Sin i/Sin r = η

In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

Thus;

Sin 30 / Sin r = 1.33

Sin r = Sin 30°/1.33

= 0.3759

r = Sin^-1 0.3759

= 22.08

<u>≈ 22°</u>

3 0

2 years ago

Two electric force vectors act on a particle. Their x-components are 13.5 N and −7.40 N and their y-components are −12.0 N and −

guapka [62]

Answer:

Explanation:

Given two vectors as follows

E₁ = 13.5 i -12 j

E₂ = -7.4 i - 4.7 j

Resultant E = E₁ + E₂

= 13.5 i -12 j -7.4 i - 4.7 j

E = 6.1 i - 16.7 j

a ) X component of resultant = 6.1 N

b ) y component of resultant = -16.7 N

Magnitude of resultant = √ ( 6.1² + 16.7² )

= 17.75 N

d ) If θ be the required angle

tanθ = 16.7 / 6.1 = 2.73

θ = 70° .

counterclockwise = 360 - 70 = 290°

6 0

2 years ago

A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which

goldfiish [28.3K]

Answer:

160 Hz , 240 Hz , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

$f_1=\dfrac{320}{4}\ Hz$

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅= 5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz , 240 Hz , 400 Hz

6 0

2 years ago

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a

Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

$y = 0.049 \cos(7t)$ .......(1)

The general equation of a wave is given by :

$y=A\cos(\omega t)$ .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0

2 years ago

Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B

gregori [183]

Answer:

The force P required is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

$tan \ \alpha = \dfrac{CD}{BD}$

where;

BD = AB-AD

Then;

$tan \ \alpha = \dfrac{CD}{AB-AD}$

$\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )$

replacing their respective values, where;

CD = 2 sin 30° m, AB = 2m and AD = 2 cos 30° m

$\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )$

$\alpha = tan^{-1} \bigg(3.73\bigg )$

$\alpha \simeq 75^0$

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

$T_{BC} = P$

Using the force equilibrium expression along the horizontal direction.

$\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0$

replacing the value of $\alpha \simeq 75^0$

$-T_{AC} \ cos 30^0 + P cos 75^0 = 0$

$P \ cos \ 75^0 = T_{AC} \ cos \ 30^0$

$P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)$

Along the vertical direction, the force equilibrium equation can be expressed as:

$\sum F_y =0$

$-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0$

$W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0$

replacing $\alpha \simeq 75^0$ and $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

Also, replacing W for (200 × 9.81) N

$200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

$200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0$

$1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)$

$1962= T_{AC} \ (0.8660\times 3.732 + 0.5)$

$1962= T_{AC} \ (3.231912 + 0.5)$

$1962= T_{AC} \ (3.731912)$

$T_{AC} = \dfrac{1962}{ \ (3.731912)}$

$T_{AC} = 525.736 \ N$

From $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \times0.866}{0.2588}$

P = 1759.22 N

Thus, the force P required is 1759.22 N

6 0

1 year ago