Shortcut Methods
Numerical 1
Shortcut Method:
- Use the formula: (B = \frac{\mu_0}{4\pi}\frac{2\pi N I}{r}) where:
- (B) is the magnetic field strength
- (\mu_0) is the vacuum permeability constant ((4\pi × 10^{−7} \ Nm^2/A^2))
- (N) is the number of turns in the coil
- (I) is the current
- (r) is the distance from the centre of the coil to the wire.
- Substitute the given values ((N=1, I=5\ A, r=20\ cm=0.2\ m)) to find (B).
Full Calculation: $$B = \frac{4\pi × 10^{-7} Nm^2/A^2}{4\pi}\frac{2\pi (1)(5\ A)}{0.2\ m}$$
$$B = 5\times 10^{-6}\ T$$
Therefore, the magnitude of the magnetic field at the centre of the coil is 5(\mu)T.
Numerical 2
Shortcut Method:
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Calculate the magnetic field produced by the current-carrying wire at the location of the loop’s centre using: (B_{wire}=\frac{\mu_0 I}{2\pi d}) where (d) is the perpendicular distance from the wire to the centre of the loop.
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Calculate the magnetic field produced by the loop itself at its own centre using: (B_{loop}=\frac{\mu_0 N I}{2R}) where (N) is the number of turns, (I) is the current, and (R) is the radius of the loop.
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Since the loop’s side is parallel to the wire, the net magnetic field at the loop’s centre will be the vector addition of (B_{wire}) and (B_{loop}).
Full Calculation: $$B_{wire} = \frac{4\pi × 10^{-7} Nm^2/A^2 (10\ A)}{2\pi (5\times10^{-2}m)}$$
$$B_{wire} = 4\times 10^{-5}\ T$$
$$B_{loop} = \frac{4\pi × 10^{-7} Nm^2/A^2 (100) (5\ A)}{2(5\times10^{-2}m)}$$
$$B_{loop} = 2\times 10^{-3}\ T$$
Net Magnetic Field: $$B_{net} = \sqrt{B_{wire}^2 + B_{loop}^2}$$
$$B_{net} = \sqrt{(4\times 10^{-5})^2 + (2\times 10^{-3})^2}$$
$$B_{net} = 2.02\times 10^{-3}\ T$$
Therefore, the magnitude of the magnetic field at the centre of the loop is 2.02(\mu)T.
-Note: The vector nature of magnetic fields is important here. B_wire and B_loop are perpendicular to each other, so we cannot simply add their magnitudes. We must use the Pythagorean theorem to find their resultant magnitude.
Numerical 3
Shortcut Method: Use the formula for the magnetic field at the centre of a circular coil: (B = \frac{\mu_0NI}{2R}) where (N) is the number of turns, (I) is the current, and (R) is the radius of the coil.
Full Calculation: $$B = \frac{(4\pi × 10^{-7} Nm^2/A^2)(20)(5\ A)}{2(5\times10^{-2} m)}$$
$$B = 4\times 10^{-3} T$$
Therefore, the magnitude of the magnetic field at the centre of the coil is 4(\mu)T.
Numerical 4
Shortcut Method: Use the formula for the magnetic field inside a solenoid: (B = \frac{\mu_0NI}{l}) where (N) is the number of turns, (I) is the current, and (l) is the length of the solenoid.
Full Calculation: $$B = \frac{4\pi × 10^{-7} Nm^2/A^2 (1000) (5\ A)}{0.2\ m}$$
$$B = 1\times 10^{-2} T$$
Therefore, the magnitude of the magnetic field inside the solenoid is 10(\mu)T.
Numerical 5
Shortcut Method: Use the formula for the magnetic field inside a toroid: (B = \frac{\mu_0NI}{2\pi r}) where (N) is the number of turns, (I) is the current, and (r) is the average radius of the toroid.
Average Radius Formula: (r_{avg} = \frac{R_1 + R_2}{2}) where (R_1) and (R_2) are the inner and outer radii.
Full Calculation: $$r_{avg} = \frac{5\ cm + 7\ cm}{2} = 6\ cm = 0.06\ m$$
$$B = \frac{4\pi × 10^{-7} Nm^2/A^2 (500)(5\ A)}{2(3.14)(0.06\ m)}$$
$$B = 1.05\times 10^{-3} T$$
Therefore, the magnitude of the magnetic field inside the toroid is 1.05(\mu)T.
