Statements:
1. What is the magnitude of force on a current-carrying conductor of length 25 cm carrying a current of 2 mA placed perpendicularly in a uniform magnetic field of 5 mT?
2. What is the magnitude of magnetic field strength in which a 15 cm conductor carrying a current of 5 mA experiences a force of 120 N?
3. A step-up transformer has a turn ratio of 2:125 and 24 V AC is applied at its primary coil. What is its secondary voltage?
4. You want to design a transformer to step down alternating voltage from 440V to 220V. What should be the number of turns of the secondary coil if the primary coil has 1000 turns?
5. A transformer steps up 115 VAC into 345 VAC. What is its turn ratio? If the secondary coil has 420 turns, find the number of turns in the primary coil.
6. When 100 VAC is applied on the primary of a step-up transformer, it gives 250 VAC as output on the secondary coil. If 30 VAC is applied on the primary coil of this transformer, then what will be the output on the secondary coil?
1. Magnitude of Force on Current-Carrying Conductor
We can use the formula for the magnetic force on a current-carrying conductor:
F=BILsin(θ)
Where:
- = force
- = magnetic field strength
- = current
- = length of the conductor
- θ= angle between the magnetic field and the conductor (since the field is perpendicular, sin(90∘)=1
Given:
- L=25 cm=0.25 m
- I=2 mA=2×10−3 A
- B=5 mT=5×10−3 T
F=(5×10−3 T)×(2×10−3 A)×(0.25 m)
Let’s calculate:
F=5×10−3×2×10−3×0.25=2.5×10−6 N
Thus, the magnitude of the force is 2.5×10−6 N
2. Magnetic Field Strength
We can rearrange the formula for magnetic force to solve for BB:
F=BIL⇒B=F/IL
Given:
- F=120 N
- I=5 mA=5×10−3 A
- L=15 cm=0.15 m
B=120 N/(5×10−3 A)×(0.15 m)
B=1207.5×10−3=1.6×104 T
So, the magnetic field strength is 1.6×104 T.
3. Secondary Voltage of Step-Up Transformer
The voltage ratio in a transformer is equal to the ratio of the number of turns in the primary and secondary coils:
Vp/Vs=Np/Ns
Where:
- Vp = primary voltage
- Vs = secondary voltage
- Np = number of turns in the primary coil
- Ns = number of turns in the secondary coil
Given:
- Vp=24 V
- Turn ratio Np/Ns=2/125
24/Vs=2/125
Solving for Vs:
Vs=24×125/2=1500 V
Thus, the secondary voltage is 1500 V.
4. Number of Turns in the Secondary Coil
Using the same formula for voltage ratio:
Vp/Vs=Np/Ns
We can solve for NsN_s:
Vp/Vs=Np/Ns ⇒ Ns=Np×Vs/Vp
Given:
- Vp=440 V
- Vs=220 V
- Np=1000 turns
Ns=1000×220/440=500 turns
So, the secondary coil should have 500 turns.
5. Turn Ratio and Number of Turns in Primary Coil
The turn ratio can be calculated using the formula:
Vp/Vs=Np/Ns
Given:
- Vp=115 V
- Vs=345 V
- Ns=420 turns
First, we calculate the turn ratio:
Vp/Vs=115/345=1/3
Thus, the turn ratio is 1/3
Now, we can calculate Np:
Np=Ns×Vp/Vs=420×115/345
So, the number of turns in the primary coil is 140 turns.
6. Output Voltage for New Primary Voltage
Using the voltage ratio formula:
Vp/Vs=Np/Ns
We can find Vs when a new primary voltage is applied:
Given:
- Vp=100 V
- Vs=250 V(initial condition)
Now, applying a primary voltage of 30 V:
100/250=30/Vs
Solving for Vs:
Vs=30×250/100=75 V
Thus, the output secondary voltage will be 75 V.