AQA A2 Physics P24 Magnetic Field Kerboodle Answers

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C24.1 Current-carrying conductors in a magnetic field AQA A2 Physics P24 Magnetic Field Kerboodle Answers: Page No. 399

1 a 2.4 x 10^-2 N; west

b 4.5 A; cast to west

c 0.20 T; vertically down

d 8.0 x 10^-3 N; due south

2 a a 22 mN due east

b b 4.0 A west to east

3 Short sides; zero, long sides; 2.72 N vertically up on one side and vertically down on the other side

4 a the magnitude of the Earth magnetic field at this position,

58 μT

b the magnitude and direction of the force on a vertical wire of length 0.90 m carrying a current of 4.5 A downwards.

6.5 x 10^-5 N due east

 

24.2 Moving charges in a magnetic field AQA A2 Physics P24 Magnetic Field Kerboodle Answers : Page No. 402

 

1 i 1.9 x 10^-13 N

ii 0

2

3.8 x 10^-23 N horizontal due East

4 a Explain why a potential difference is created across the semiconductor as a result of the application of the magnetic field.

b When the magnetic flux density was 90mT, each electron moving through the slice experiences a force Of 6.4 x N due the magnetic field.

i 4.4ms^-1

ii 8.5 x 10^-20 N

 

24.3 Charged particles in circular orbits AQA A2 Physics P24 Magnetic Field Kerboodle Answers : Page No. 405

 

1 a i Explain why the electrons move on a circular orbit. Calculate the radius of the orbit.

b 2.8 mT

2 a 4.7mT

b 17.5 mm

3 a

b 1.2 MeV

4 a

8.0 x 10⁶ c kg^-1

b 1.4 x 10⁷ c kg^-1

 

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Practice Questions: Page No. 406-409

 

1 (a)

The units of the quantities are:

Force F: newton (N)

Current I: ampere (A)

Magnetic flux density B: tesla (T) or weber metre^-2 (Wb m^-2)

Length l of wire in field: metre (m)

The equation F = BIl applies only when the magnetic field is directed at right angles to the direction of the current.

(b)

 (i) density of copper 8.9 x 10³kgm^-3.

Mass m of bar

= (25 × 10^-3)² × 8900 × l

= 5.56 l

Weight mg of bar = 5.56l × 9.81 = 54.6l

Magnetic force = weight of bar

∴ mg = BIl

BIl = 54.6l gives

(ii)

2

(a) (i)

(ii)

Force on X due to Y:

• B at X (due to Y) is unchanged because the current in Y is unaltered
• F = BIl and the current in X is halved, so the force on X is now 90 N.

Force on Y due to X:

• B at Y (due to X) is halved because the current in X is halved
• F = BIl and the current in Y is unchanged, so the force on Y is also now 90 N.

(b) (i)

• The magnetic field due to the current in each bus bar alternates with the current in the bar
• The force will be zero when the current in either wire is zero
• Both current and field reverse together, so the force does not reverse direction when the current reverses
• The force on each wire varies periodically, making the wires vibrate
• The frequency of vibration is twice the ac frequency.

(ii)

The amplitude of vibration could be reduced by:

• clamping each bar along it length
• attaching a damping device to each bar
• moving the bars further apart.

3 (a)

 (i)

The force on the ion is directed out of the plane of the page.

(ii)

In a magnetic field the ion moves in a circular path…in a horizontal plane (or out of the plane of the page).

The force equation for the circular motion is

(b) (i) The strength of the magnetic field is doubled.

(ii)

When magnetic field strength is doubled

• the radius of the path decreases
• to half the original value.

When a single charged ion is used

• the radius of the path increases
• to double the original value.

4

(a) (i)

The flux density of a magnetic field is 1 tesla if a 1 metre length of wire carrying a current of 1 ampere experiences a magnetic force of 1 newton when placed in the field…in such a way that the magnetic field is at right angles to the current.

(ii)

Within the velocity selector, magnetic force on

ion = electric force on ion

∴ BQv = EQ

In this equation Q cancels, giving

Bv = E, and velocity of ions v =E/V

meaning that v does not depend on Q.

(iii)

(b) (i)

5

(a)

(b)

Since the particles follow the same path, the radius of curvature r is the same.

They have the same momentum because r ∝ mv when B and Q are the same. mv is the same, but m is different for the antiproton and the negative pion,

∴ v must be different.

(c)

Identify correct format for both particles:

antiproton: 3 antiquarks

negative pion: quark + antiquark

antiproton: 2 up antiquarks +1 down antiquark ( u , u , d )

negative pion: 1 up antiquark + 1 down quark ( u , d )

The magnetic field is directed into the plane of the page

(ii)

The magnetic field is directed at right angles to the velocity of the ions.

• The magnetic force acts at right angles to both the magnetic field and the velocity.
• Hence the magnetic force acts at right angle to the velocity of the ions.
• This force changes the direction of the velocity of the ions but not its magnitude.
• The force remains perpendicular to the velocity as the direction of movement of the ions changes…
• so the force acts as a centripetal force.

(iii)

(b)

(c)

1. i)

Relevant points include:

(i) Ions have a different mass.

Diameter d of path ∝ mass m of ion.

Due to isotopes of the same element.

(ii) Ions are doubly ionised.

Diameter d of path ∝ 1/Q; if Q is doubled then d is halved.

7  (a)

 

(b)

As the kinetic energy of the protons increases, the magnetic flux density has to increase…because in a path of constant radius (using the same charged particles) magnetic flux

density B ∝ v

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