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Chapter 13: Problem 1972

A magnet of magnetic moment \(50 \uparrow \mathrm{A} \mathrm{m}^{2}\) is placed along the \(\mathrm{X}\) -axis in a mag. field $\mathrm{B}^{-}=(0.5 \uparrow+3.0 \mathrm{~J} \wedge$ ) Tesla. The torque acting on the magnet is N.m. (c) \(75 \mathrm{k} \wedge\) (d) \(25 \sqrt{5} \mathrm{k} \wedge\) (a) \(175 \mathrm{k}\) (b) \(150 \mathrm{k}\)

Short Answer

Expert verified
The torque acting on the magnet is \(\boldsymbol{\tau} = 150 \hat{k}\), which corresponds to answer (b).

Step by step solution

01

Write down the given magnetic moment and magnetic field as vectors

The magnetic moment is given as \(50 \hat{i}\), which in vector form can be written as \(\boldsymbol{m} = 50 \hat{i} = (50, 0, 0)\). The magnetic field is given as \(0.5 \hat{i} + 3.0 \hat{j}\), which in vector form can be written as \(\boldsymbol{B} = 0.5 \hat{i} + 3.0 \hat{j} = (0.5, 3.0, 0)\).
02

Calculate the torque by finding the cross product of the magnetic moment and magnetic field

The cross product of two vectors \((a_x, a_y, a_z)\) and \((b_x, b_y, b_z)\) is given by: \[ (a_x, a_y, a_z) \times (b_x, b_y, b_z) = (a_yb_z - a_zb_y, a_zb_x - a_xb_z, a_xb_y - a_yb_x) \] Applying this formula to the magnetic moment and magnetic field vectors, we get: \[ \boldsymbol{\tau} = (50, 0, 0) \times (0.5, 3.0, 0) = (0, 0, 50 \times 3.0 - 0 \times 0.5) = (0, 0, 150) \] So, the torque acting on the magnet is \(\boldsymbol{\tau} = 150 \hat{k} \), which corresponds to answer (b).

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Most popular questions from this chapter

A magnetic field \(B^{-}=\) Bo \(j \wedge\) exists in the region \(a

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