Magnetic Flux through an Aluminum Window
An aluminum window has a width of 60 cm and length of 85 cm. When the window is closed, a magnetic flux density of 1.8 × 10-4 T is normal to the window. Calculate the magnetic flux through the window.
Given Data:
- Width of Window (w) = 60 cm = 0.60 m
- Length of Window (L) = 85 cm = 0.85 m
- Magnetic flux density (B) = 1.8 × 10-4 T
- Angle (θ) = 0° (Since field is normal to the window, the angle with the area vector is 0°)
- Magnetic Flux (Φ) = ?
Solution:
The Magnetic Flux is given by:
First, calculate the Area of the window (A):
Putting the values into equation (i):
Result:
The magnetic flux through the window is 9.18 × 10-5 Wb.
Flux Between Horseshoe Magnet Poles
The poles of a horse shoe magnet measure 8 cm × 3.2 cm. The magnetic flux density between the magnet poles is 80 mT. Outside of the magnet, the magnetic flux density is zero. Calculate the magnetic flux between the poles of the magnet.
Given Data:
- Dimensions of Pole face = 8 cm × 3.2 cm = 0.08 m × 0.032 m
- Area of Magnet (A) = 0.08 × 0.032 = 2.56 × 10-3 m2
- Magnetic flux density (B) = 80 mT = 80 × 10-3 T = 0.08 T
- Magnetic Flux (Φ) = ?
Solution:
The Magnetic Flux is given by:
Substituting the values:
Result:
The magnetic flux between the poles of the magnet is 2.05 × 10-4 Wb.
Magnetic Force on a Current-Carrying Wire
A wire 1.80 m long carries a current of 13.0 A and makes an angle of 35.0° with a uniform magnetic field of magnitude B = 1.50 T. Calculate the magnetic force on the wire.
Given Data:
- Length of Conductor (L) = 1.80 m
- Electric Current (I) = 13.0 A
- Angle (θ) = 35.0°
- Magnetic Field (B) = 1.50 T
- Magnetic Force (F) = ?
Solution:
The Force on a Current-Carrying Conductor in a magnetic field is given by:
Substituting the given values:
Result:
The magnetic force on the wire is 20.13 N (approximately 20 N).
Magnetic Field at the Center of a Solenoid
A solenoid has length L = 1.23 m and inner diameter d = 3.55 cm, and it carries a current I = 5.57 A. It consists of five close-packed layers, each with 850 turns along length L. What is B at its center?
Given Data:
- Length of Solenoid (L) = 1.23 m
- Diameter (d) = 3.55 cm = 0.0355 m (Extra data, not required for calculation)
- Electric Current (I) = 5.57 A
- Turns per layer = 850 turns
- Total Layers = 5
- Total Number of Turns (N) = 5 × 850 = 4250 turns
- Permeability of free space (μ0) = 4π × 10-7 T·m/A
- Magnetic Field (B) = ?
Solution:
The magnetic field inside a long solenoid is given by:
Where total turns per unit length (n) is:
Putting values in equation (i):
Result:
The Magnetic Field B at its center is 2.42 × 10-2 T.
Conversion of Galvanometer into Voltmeter
A moving coil galvanometer has a resistance of 50 ohms and it gives full scale deflection at 4mA current. A voltmeter is made using this galvanometer and a 5 kΩ resistance. Calculate the maximum voltage that can be measured using this voltmeter.
Given Data:
- Resistance of Galvanometer (Rg) = 50 Ω
- Full Scale Deflection Current (Ig) = 4 mA = 4 × 10-3 A = 0.004 A
- Series High Resistance (Rx) = 5 kΩ = 5000 Ω
- Maximum Voltage range (V) = ?
Solution:
The formulation for the high multiplier series resistance of a voltmeter is:
Isolating the total voltage equation:
Substituting the values:
Result:
The maximum voltage that can be measured using this voltmeter is 20.2 V.
Magnetic Field of a Long Wire vs. Earth's Field
Compute the magnitude of the magnetic field of a long, straight wire carrying a current of 1 A at distance of 1 m from it. Compare it with Earth's magnetic field ($B_e = 5.0 \times 10^{-5}$ T).
Given Data:
- Electric Current (I) = 1 A
- Distance (r) = 1 m
- Earth's Magnetic Field (Be) = 5.0 × 10-5 T
- Magnetic Field of Wire (B) = ?
Solution:
The magnetic field due to a long current-carrying straight conductor is:
Substituting values where $\mu_0 = 4\pi \times 10^{-7}$ T·m/A:
Now, comparing it with Earth's magnetic field ratio:
Result:
The magnitude of the magnetic field is 2 × 10-7 T, which is 0.004 times (or 1/250th) the strength of Earth's magnetic field.
Finding Current for Specific Field Strength
Find the current in a long straight wire that would produce a magnetic field twice the strength of the Earth's at a distance of 5.0 cm from the wire. (Magnetic field of Earth = 5.0 × 10-5 T).
Given Data:
- Earth's Magnetic Field (Be) = 5.0 × 10-5 T
- Required Field Strength (B) = 2 × Be = 2 × (5.0 × 10-5 T) = 1.0 × 10-4 T
- Distance (r) = 5.0 cm = 0.05 m
- Electric Current (I) = ?
Solution:
Magnetic field formula for a straight line conductor:
Rearranging the equation to solve explicitly for electric current (I):
Substituting values:
Result:
The current in the long straight wire is 25 A.
Flux Density and Mutual Force Per Unit Length
What is the flux density at a distance of 0.1 m in air from a long straight conductor carrying a current of 6.5 A? Calculate the force per unit length on a similar parallel conductor at a distance of 0.1 m from the first and carrying a current of 3 A.
Given Data:
- Distance (r) = 0.1 m
- Current in 1st wire (I1) = 6.5 A
- Current in 2nd parallel wire (I2) = 3 A
- Magnetic Flux Density (B) = ?
- Force per unit length (F/L) = ?
Solution:
Part 1: Calculate magnetic field/flux density due to the first wire:
Part 2: Calculate the magnetic force per unit length on the second wire:
Result:
The flux density at a distance of 0.1 m in air is 1.3 × 10-5 T and the force per unit length on the second wire is 3.9 × 10-5 N/m.
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