Does an electric current create a magnetic field ?

A loop of current-carrying wire has a magnetic dipole moment of 5. 0 10–4 am2. if the dipole moment makes an angle of 57° with a

Digiron [165]

The potential energy will be 1.46*10^-4J.

To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.

<h3>How to find the potential energy of the loop?</h3>

We have the expression for torque acting on a current loop in a uniform magnetic field as,

$\tau=MBsin\theta$

where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.

As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
Thus, the potential energy will be,

$\tau=W=U=MBsin\theta=5*10^{-4}*0.35*sin57=1.46*10^{-4}J$

Thus, we can conclude that, the potential energy will be 1.46*10^-4J.

Learn more about the torque here:

brainly.com/question/27949876

#SPJ4

7 0

2 years ago

There are no risks when taking medication to treat anxiety.

lidiya [134]

Answer:

False

Explanation:

5 0

3 years ago

Read 2 more answers

shirley knott, a 65kg astronaut holds a 12kg tank of oxygen, which she throws backwards at 2.2m/s. with what speed does she move

babymother [125]

This question involves the concepts of the law of conservation of momentum and speed.

She moves with a speed of "0.41 m/s".

Applying the law of conservation of momentum to this situation, we get:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = mass of shirley = 65 kg

m₂ = mass of tank = 12 kg

u₁ = initial speed of shirley = 0 m/s

u₂ = initial speed of tank = 0 m/s

v₁ = final speed of shirley = ?

v₂ = final speed of tank = 2.2 m/s

Therefore,

$(65\ kg)(0\ m/s)+(12\ kg)(0\ m/s)=(65\ kg)(v_1)+(12\ kg)(2.2\ m/s)\\\\v_1=\frac{-26.4\ kg.m/s}{65\ kg}\\\\v_1=-0.41\ m/s$

negative sign shows the opposite direction.

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

3 0

2 years ago

White light containing wavelengths from 410 nm to 750 nm falls on a grating with 7800 slits/cm. Part APart complete How wide is

Free_Kalibri [48]

Answer:

1.227 m

Explanation:

Given that

Minimum wavelength is 410 nm

Maximum wavelength is 750 nm

Grating is 7800 slits/cm

Distance is 3.2 m

To solve this question, we would use the formula

sin θ = λ/d

sin θ = (410*10^-9) / (0.01/7800)

Sin θ = 410*10^-9 / 1.282*10^-6

Sin θ = 0.32 and θ = 18.67 degrees

For the second wavelength = 750 nm

sin θ = [(0.32x750)/410]

sin θ = (240 / 410)

sin θ = 0.5853 or

θ = 35.8 degrees

And finally, the width of spectrum would be

3.2[tan 35.8 - tan 18.67]

3.2 * 0.3833

= 1.227 m

4 0

4 years ago

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0

4 years ago