Numerical 1

Shortcut Method:

1. 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.

2. Substitute the given values ((N=1, I=5\ A, r=20\ cm=0.2\ m)) to find (B).

Full Calculation: $$B=\frac{4\pi \times {10}^{-7}N{m}^2/A^2}{4\pi }\frac{2\pi (1)(5A)}{0.2m}$$

$$B=5\times {10}^{-6}T$$

Therefore, the magnitude of the magnetic field at the centre of the coil is 5(\mu)T.

Numerical 2

Shortcut Method:

1. 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.

2. 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.

3. 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: $$B_{wire}=\frac{4\pi \times {10}^{-7}N{m}^2/A^2(10A)}{2\pi (5\times {10}^{-2}m)}$$

$$B_{wire}=4\times {10}^{-5}T$$

$$B_{loop}=\frac{4\pi \times {10}^{-7}N{m}^2/A^2(100)(5A)}{2(5\times {10}^{-2}m)}$$

$$B_{loop}=2\times {10}^{-3}T$$

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: $$B=\frac{(4\pi \times {10}^{-7}N{m}^2/A^2)(20)(5A)}{2(5\times {10}^{-2}m)}$$

$$B=4\times {10}^{-3}T$$

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: $$B=\frac{4\pi \times {10}^{-7}N{m}^2/A^2(1000)(5A)}{0.2m}$$

$$B=1\times {10}^{-2}T$$

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: $$r_{avg}=\frac{5cm+7cm}{2}=6cm=0.06m$$

$$B=\frac{4\pi \times {10}^{-7}N{m}^2/A^2(500)(5A)}{2(3.14)(0.06m)}$$

$$B=1.05\times {10}^{-3}T$$

Therefore, the magnitude of the magnetic field inside the toroid is 1.05(\mu)T.