The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 andat steam point is 5.23 Ω. When it is inserted in a hot bath, the resistance of the wire is 5.795 Ω.Calculate the temperature of the bath.18. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 Texperiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of magneticmoment of the magnet?19. A 60-mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of thecurrent in the circuit.20. Explain hard ferromagnetic and soft ferromagnetic materials.21. Two magnetic poles, one of which is four times stronger than the other, exert a force of 5gf on each other whenplaced at a distance of 10 cm. Find the strength of each pole
Question
The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 andat steam point is 5.23 Ω. When it is inserted in a hot bath, the resistance of the wire is 5.795 Ω.Calculate the temperature of the bath.18. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 Texperiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of magneticmoment of the magnet?19. A 60-mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of thecurrent in the circuit.20. Explain hard ferromagnetic and soft ferromagnetic materials.21. Two magnetic poles, one of which is four times stronger than the other, exert a force of 5gf on each other whenplaced at a distance of 10 cm. Find the strength of each pole
Solution
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To calculate the temperature of the bath, we first need to find the resistance ratio. The resistance ratio (R) is the resistance at the temperature T (RT) divided by the resistance at 0°C (R0). So, R = RT/R0 = 5.795/5 = 1.159. The temperature can be calculated using the Callendar-Van Dusen equation for platinum resistance thermometers: T = -100 + sqrt(3900 + 100*(R - 1)) = -100 + sqrt(3900 + 100*(1.159 - 1)) = 100°C.
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The torque (τ) on a magnet in a uniform magnetic field is given by τ = mBsin(θ), where m is the magnetic moment, B is the magnetic field strength, and θ is the angle between the magnetic moment and the magnetic field. Rearranging for m gives m = τ / (Bsin(θ)) = 4.510^-2 / (0.25*sin(30)) = 0.36 J/T.
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The rms current (I) in an AC circuit with a capacitor is given by I = V / Xc, where V is the rms voltage and Xc is the capacitive reactance. Xc = 1 / (2pifC), where f is the frequency and C is the capacitance. So, I = 110 / (1 / (2pi6060*10^-6)) = 0.029 A.
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Hard ferromagnetic materials are those that retain their magnetization and are hard to demagnetize. They are used to make permanent magnets. Soft ferromagnetic materials, on the other hand, can be easily magnetized and demagnetized, and are used in applications like transformer cores.
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The force (F) between two magnetic poles is given by F = kq1q2 / r^2, where k is the magnetic constant, q1 and q2 are the pole strengths, and r is the distance between the poles. If q2 = 4q1, then F = kq14q1 / r^2. Solving for q1 gives q1 = sqrt(Fr^2 / (4k)) = sqrt(510^-30.1^2 / (410^-7)) = 0.025 N/Wb, and q2 = 4q1 = 0.1 N/Wb.
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