MAGNETIC FIELDS
MAGNETIC FORCES
An alpha particle with a speed of \( 4.4 \times 10^5 m/s \) moves perpendicular to a \( 0.75T \) magnetic field in a circular path with a radius of \( 0.012m \). What is the magnitude of the charge on the helium nucleus?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
INFORMATION GIVEN
- A helium nucleus has a mass of \( 6.6 \times 10^{-27} kg \)
- The speed of the alpha particle is \( 4.4 \times 10^5 m/s \)
- \( r = 0.012m \)
- \( B = 0.75T \)
USEFUL FORMULA(S)
- \( F_{c} = F_{m} \)
- \( \frac{mv^2}{r} = qvB_{\bot} \)
- \( q = \frac{mv}{rB} \)
PROCESS
- Using our formula, plug in our known values:
- \( q = \frac{mw}{rB} \)
- \( q = \frac{(6.6 \times 10^{-27} kg)(4.4 \times 10^5 m/s)}{(0.012m)(0.75T)} \)
- \( \Rightarrow q = 3.2 \times 10^{-19}C \)
- The magnitude of the charge on the helium nucleus is \( 3.2 \times 10^{-19} C \)
CATHODE RAY TUBES
In a cathode ray tube electrons are accelerated from rest by a potential difference of \( 1.40kV \). If these electrons enter a magnetic field with a strength of \( 0.0220T \), what is the radius of arc for the deflected electrons?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
INFORMATION GIVEN
- \( V = 1.40 \times 10^3 V \)
- \( B_{\bot} = 0.0220T \)
- \( q = 1.60 \times 10^{-19} C \)
- \( m = 9.11 \times 10^{-31} kg \)
USEFUL FORMULA(S)
- \( F_{c} = F_{m} \)
- \( \frac{mv^2}{r} = qvB_{\bot} \)
- \( r = \frac{mv}{qB_{\bot}} \)
- \( E_{k} = qV \)
- \( \frac{1}{2}mv^2 = qV \)
- \( v = \sqrt{ \frac{2qV}{m} } \)
PROCESS
- First, we need to solve for \( v \).
- \( v = \sqrt{ \frac{2qV}{m} \)
- \( v = \sqrt{ \frac{2(1.60 \times 10^{-19} C)(1400V)}{9.11 \times 10^{-31}kg} } \)
- \( \Rightarrow v = 2.22x10^7m/s \)
- Then, we use that value to solve for \( r \).
- \( r = \frac{mv}{qB_{\bot}} \)
- \( r = \frac{(9.11 \times 10^{-31}kg)(2.22 \times 10^7m/s)}{(1.60 \times 10^{-19}C)(0.0220T)} \)
- \( \Rightarrrow r = 5.74 \times 10^{-3}m \)
- The radius of the arc for the deflected electrons is \( 5.74 \times 10^{-3} m \).
MOTORS
A \( 12cm \) long wire hangs perpendicularly across a magnetic field whose density is \( 0.015T \). How great must the current be to produce a motor effect of \( 1.0N \)?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
INFORMATION GIVEN
- \( F_{m} = 1.0N \)
- \( B_{\bot} = 0.015T \)
- \( L = 0.12m \)
USEFUL FORMULA(S)
- \( F_{m} = B_{\bot}IL \)
- \( I = \frac{F_{m}}{B_{\bot}L} \)
PROCESS
- Using our formula, plug in our known values:
- \( I = \frac{F_{m}}{B_{\bot}L} \)
- \( I = \frac{1.0N}{(0.015T)(0.12m)} \)
- \( \Rightarrow I = 5.6 \times 10^2A \)
- The current must have a strength of \( 5.6 \times 10^2 A \) to produce a motor effect of \( 1.0 N \).
TRANSFORMERS
An ideal transformer has 40 turns on its secondary coil and 3200 turns on its primary coil.
PROCESS
- Calculate the secondary voltage if a 220V a.c. supply is connected to the primary coil
- Calculate the current through a 15 Ohm resistor connected to the secondary coil.
- What current must flow through the primary coil when the 15 Ohm resistor is connected to the secondary coil?
- What would be the answer to question 1 be if a d.c. supply was used instead?
- Primary turns: 3200
- Secondary turns: 40
- Primary voltage: 220V
- \( V_{s} = V_{p} x (\frac{n_{s}}{n_{p}}) \)
- \( V = IR \) so \( I = \frac{V}{R} \)
- \( P = VI \)
PROCESS
- \( V_{s} = V_{p} x (\frac{n_{s}}{n_{p}}) \), so \( V_{s} = 220 \cdot \frac{40}{3200} = 220 \cdot 0.0125 = 2.75V \)
- \( I = \frac{V}{R} \), so \( I = \frac{2.75}{15} = 0.18A \)
- \( P = VI \), so the power in the secondary coil is \( 2.75V \cdot 0.18A = 0.5W \). The power in the primary coil must be the same, so the current flowing through the primary coil is \( \frac{P}{V} = \frac{0.5}{220} = 0.002A \).
- As transformers do not work with a d.c. power supply, the answer would be \( 0V \). You need a changing magnetic field in order to induce an electric current in the secondary coil.