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1. A proton with charge 1.602 x 10^-19 C moves at a speed of 300 m/s in a magnetic field at an angle of 65 degrees. If the strength of the magnetic field is 19 T, what would be the magnitude of the force the charge experiences?

2.Where are the magnetic field lines of a permanent magnet the strongest?

3,Look at the picture of a positive charge moving in a magnetic field. Using the right hand rule, which direction will the force be that the charge experiences?
4.An alpha particle travelling at 2155 m/s enters a magnetic field of strength 12.2 T. The particle is moving horizontally and the magnetic field is vertical. If an alpha particle contains two protons, each with a charge of 1.602 x 10^-19 C and the particle has a mass of 6.64 x 10^-22 kg, what is the radius of the circular path the particle will travel in?
5.What is the cyclotron frequency of an electron entering a magnetic field of strength 0.0045 T? The charge of an electron is -1.602 x 10^-19 C and the mass of an electron is 9.31 x 10^-31 kg
6,If a charged particle is travelling in a helical shape as it moves through a magnetic field, but then the particle gains the opposite charge, what happens to it's travelling path?
7.A conducting loop is placed in a magnetic field. What must be true for there to be a current induced in the loop?
8.A rectanglular loop of length 15 cm and width 8 cm is placed in a horizontal plane. A magnetic field of strength 5.5 T passes through the plane at 18 degrees above the horizontal. What is the flux through the loop? (1 point)
9.A conducting coil with 100 loops is placed in a magnetic field. The radius of each loop is 0.075 m. The magnetic field passes through the coil at an angle of 60 degrees. If the magnetic field increases at a rate of 0.250 T/s, what is the emf produced in the coil after 1 second? (1 point)
10. A transformer coil has 20 turns on one end and 200 turns on the other end. An emf of 300 V comes into the 20 turn end. How much emf comes out of the 200 turn end of the transformer? (1 point)

Respuesta :

Answer:

1)

[tex]F = 8.28 \times 10^{-16} N[/tex]

2)

Magnetic field near to poles of the magnet is strongest in its strength

3)

with the help of right hand rule we can find the direction of the force

4)

[tex]R = 3.66 \times 10^{-6} m[/tex]

5)

[tex]f = 1.23 \times 10^8 Hz[/tex]

6)

Axis of the helical path will change but the path will be same as initial

7)

[tex]EMF = \frac{d\phi}{dt}[/tex]

8)

[tex]\phi = 0.02 Wb[/tex]

9)

[tex]EMF = 0.22 V[/tex]

10)

[tex]V = 3000 V[/tex]

Explanation:

1)

Magnetic force is given as

[tex]F = qvBsin\theta[/tex]

so we have

[tex]F = (1.602 \times 10^{-19})(300)(19)sin65[/tex]

[tex]F = 8.28 \times 10^{-16} N[/tex]

2)

Magnetic field near to poles of the magnet is strongest in its strength

3)

with the help of right hand rule we can find the direction of the force

put the fingers of right hand along the velocity of charge and curl is towards the magnetic field then the direction of thumb will show the direction of force on positive charge moving in magnetic field

4)

Radius of the charge moving in magnetic field is given as

[tex]R = \frac{mv}{qB}[/tex]

so we have

[tex]R = \frac{6.64 \times 10^{-27} \times 2155}{(2\times 1.602 \times 10^{-19}) 12.2}[/tex]

[tex]R = 3.66 \times 10^{-6} m[/tex]

5)

[tex]f = \frac{qB}{2\pi m}[/tex]

so we have

[tex]f = \frac{(1.602 \times 10^{-19})(0.0045)}{2\pi(9.31 \times 10^{-31})}[/tex]

[tex]f = 1.23 \times 10^8 Hz[/tex]

6)

Axis of the helical path will change but the path will be same as initial

So the new path will be same as helical path but the axis of the helical path will change

7)

When magnetic flux linked with the loop will change with time then as per Farady's law the induced EMF is given as

[tex]EMF = \frac{d\phi}{dt}[/tex]

8)

magnetic flux is given as

[tex]\phi = BA cos\theta[/tex]

so we have

[tex]\phi = (5.5)(0.15\times 0.08) sin18[/tex]

[tex]\phi = 0.02 Wb[/tex]

9)

EMF as per faraday's law is given as

[tex]EMF = NA\frac{dB}{dt} cos\theta[/tex]

so we have

[tex]EMF = (100)(\pi \times 0.075^2)(0.250) cos60[/tex]

[tex]EMF = 0.22 V[/tex]

10)

As per the law of transformer we know that

[tex]\frac{E_1}{E_2} = \frac{N_1}{N_2}[/tex]

so we have

[tex]\frac{300}{V} = \frac{20}{200}[/tex]

so we have

[tex]V = 3000 V[/tex]

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