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Chapter 13: Problem 1963

A magnet of length \(0.1 \mathrm{~m}\) and pole strength $10^{-4} \mathrm{~A} \mathrm{~m} .$ is kept in a magnetic field of 30 tesla at an angle of 300 . The Torque acting on it is \(\times 10^{-4}\) Joule. (a) \(7.5\) (b) 3 (c) \(1.5\) (d) 6

Short Answer

Expert verified
The correct answer is (c) \(1.5 * 10^{-4}\) Joule.

Step by step solution

01

Calculate magnetic moment

To calculate the magnetic moment (m), we use the formula: Magnetic moment (m) = pole strength (p) * length (l) where p = \(10^{-4}\) A·m and l = 0.1 m. m = \(10^{-4}\;A·m * 0.1\;m\) m = \(10^{-5} A·m^2\)
02

Calculate the torque

Now, we need to find the torque (τ) acting on the magnet using the formula: Torque (τ) = magnetic moment (m) * magnetic field (B) * sin(θ) where m = \(10^{-5}\;A·m^2\), B = 30 T, and θ = 30°. First, we need to convert the angle to radians: θ (in radians) = (30° * π) / 180° θ = π/6 Now, we can calculate the torque: τ = \(10^{-5}\;A·m^2 * 30\;T * sin(π/6)\) τ = \(3 * 10^{-4}\;N·m * 0.5\) τ = \(1.5 * 10^{-4}\;N·m\)
03

Compare the torque with the options

We have found that the torque acting on the magnet is \(1.5 * 10^{-4}\;N·m\). Comparing this value with the given options: (a) 7.5 * \(10^{-4}\) Joule (b) 3 * \(10^{-4}\) Joule (c) 1.5 * \(10^{-4}\) Joule (d) 6 * \(10^{-4}\) Joule So the correct answer is (c) 1.5 * \(10^{-4}\) Joule.

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