Question
A magnet of magnetic moment $10 \hat{i} \mathrm{~A}-\mathrm{m}^{2}$ is placed along $x$ -axis in a magnetic field $(\hat{i}+2 \hat{j})$ tesla. The torque acting on the magnet is(a) $20 \mathrm{~N}-\mathrm{m}$ along $x$ -axis(b) $10 \sqrt{5} \mathrm{~N}-\mathrm{m}$ along $z$ -axis(c) $20 \mathrm{~N}-\mathrm{m}$ along $z$ -axis(d) $10 \sqrt{5} \mathrm{~N}-\mathrm{m}$ along $y$ -axis
Step 1
Step 1: The torque acting on a magnet in a magnetic field is given by the formula $\tau = \vec{M} \times \vec{B}$, where $\vec{M}$ is the magnetic moment of the magnet and $\vec{B}$ is the magnetic field. Show more…
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A bar magnet has a magnetic moment equal to $5 \times 10^{-5} \mathrm{~Wb}-\mathrm{m} .$ It is suspended in a magnetic field which has a magnetic induetion $B$ equal to $8 \pi \times 10^{-4} \mathrm{~T}$. The magnet vibrates with a period of vibration equal to 15 s. The moment of inertia of magnet is (a) $4.54 \times 10^{4} \mathrm{~kg}-\mathrm{m}^{2}$ (b) $4.54 \times 10^{-5} \mathrm{~kg}-\mathrm{m}^{2}$ (c) $4.54 \times 10^{-4} \mathrm{~kg}-\mathrm{m}^{2}$ (d) $4.54 \times 10^{5} \mathrm{~kg}-\mathrm{m}^{2}$
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Round 2
A bar magnet of length $10 \mathrm{~cm}$ and having pole strength equal to $10^{-3} \mathrm{~Wb}$ is kept in a magnetic field having magnetie induction $B$ equal to $4 \pi \times 10^{-a} T$ It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is (a) $0.5 \mathrm{Nm}$ (b) $2 \pi \times 10^{-5} \mathrm{Nm}$ (c) $\pi \times 10^{-5} \mathrm{Nm}$ (d) $0.5 \times 10^{-5} \mathrm{Nm}$
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