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Chapter 13: Problem 1965

A small bar magnet of moment \(\mathrm{M}\) is placed in a uniform field of \(\mathrm{H}\). If magnet makes an angle of \(30^{\circ}\) with field, the torque acting on the magnet is (a) \(\mathrm{MH}\) (b) \((\mathrm{MH} / 2)\) (c) \((\mathrm{MH} / 3)\) (d) \((\mathrm{MH} / 4)\)

Short Answer

Expert verified
The torque acting on the magnet is \(\frac{MH}{2}\).

Step by step solution

01

Identify the given values.

The information given is: - Magnetic moment, M - Magnetic field, H - Angle between the magnet and the field, θ = 30°
02

Apply the formula for torque.

The formula for torque acting on the magnetic dipole in a magnetic field is: τ = MH sin(θ) Plug in the given values: τ = M * H * sin(30°)
03

Calculate the sin value.

To find the torque, we first need to calculate the value of sin(30°). Using the sine function: sin(30°) = 1 / 2
04

Calculate the torque.

Now, substitute the sin value back into the equation: τ = M * H * (1 / 2) Simplify the equation: τ = (MH) / 2
05

Compare the result with the given options.

We have found that the torque acting on the magnet is (MH) / 2. This matches the option (b). Therefore, the correct answer is: (b) (MH / 2)

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Most popular questions from this chapter

The angles of dip at two places are \(30^{\circ}\) and \(45^{\circ}\). The ratio of horizontal components of earth's magnetic field at the two places will be (a) \(\sqrt{3}: \sqrt{2}\) (b) \(1: \sqrt{2}\) (c) \(1: 2\) (d) \(1: \sqrt{3}\)

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