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A short magnetic needle is placed in a magnetic field $B \hat{i}$ in the direction $(\sqrt{3} \hat{i}+\hat{j})$.
The needle experiences a torque of $0.06 \mathrm{~N}-\mathrm{m}$. If the same magnetic needle is placed in a magnetic field $2 B \hat{j}$ in the direction $(\hat{i}+\sqrt{3} \hat{j})$, the torque experienced by it is
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The needle experiences a torque of $0.06 \mathrm{~N}-\mathrm{m}$. If the same magnetic needle is placed in a magnetic field $2 B \hat{j}$ in the direction $(\hat{i}+\sqrt{3} \hat{j})$, the torque experienced by it is
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Verified Answer
The correct answer is:
$0.12 \mathrm{~N}-\mathrm{m}$
( ) Torque on a magnet (dipole moment $\mathbf{m}$ ) when it is placed in region of magnetic field (Intensity B) is $\tau=\mathbf{M} \times \mathbf{B}$
Where, $\mathbf{M}=\mathbf{M}$. $\hat{\mathbf{u}}$
$M=$ magnitude of dipole moment and $\hat{\mathbf{u}}=$ unit vector in direction of $\mathbf{M}$ Now in case I given,
$$
\mathbf{B}=B \hat{i} \text { and } \mathbf{M}=\frac{M}{2}(\sqrt{3 i}+\hat{j})
$$
Hence, torque is $\tau_1=\frac{M}{2}(\sqrt{3} \hat{i}+\hat{j}) \times B \hat{i}$
$$
=\frac{M B}{2}(\hat{j} \times \hat{i})=\frac{M B}{2}(-\hat{k})
$$
Now, given, $\left|\tau_1\right|=0.06 \mathrm{~N}-\mathrm{m}$
So,
$$
0.06=M B / 2 \text { or } M B=0.12 \text { units }
$$
Now, when $\mathbf{M}=M\left(\frac{\hat{i}+\sqrt{3} \hat{j}}{2}\right)$
and $\mathbf{B}=2 \hat{B j}$ then, torque will be $\tau_2=\mathbf{M} \times \mathbf{B}$
$$
=\frac{M}{2}(\hat{i}+\sqrt{3} \hat{j}) \times 2 B \hat{j}=M B(\hat{i} \times \hat{j})=M B(\hat{k})
$$
$\therefore$ Magnitude of torque is $\left|\tau_2\right|=M B=012 \mathrm{~N}-\mathrm{m}$
Where, $\mathbf{M}=\mathbf{M}$. $\hat{\mathbf{u}}$
$M=$ magnitude of dipole moment and $\hat{\mathbf{u}}=$ unit vector in direction of $\mathbf{M}$ Now in case I given,
$$
\mathbf{B}=B \hat{i} \text { and } \mathbf{M}=\frac{M}{2}(\sqrt{3 i}+\hat{j})
$$
Hence, torque is $\tau_1=\frac{M}{2}(\sqrt{3} \hat{i}+\hat{j}) \times B \hat{i}$
$$
=\frac{M B}{2}(\hat{j} \times \hat{i})=\frac{M B}{2}(-\hat{k})
$$
Now, given, $\left|\tau_1\right|=0.06 \mathrm{~N}-\mathrm{m}$
So,
$$
0.06=M B / 2 \text { or } M B=0.12 \text { units }
$$
Now, when $\mathbf{M}=M\left(\frac{\hat{i}+\sqrt{3} \hat{j}}{2}\right)$
and $\mathbf{B}=2 \hat{B j}$ then, torque will be $\tau_2=\mathbf{M} \times \mathbf{B}$
$$
=\frac{M}{2}(\hat{i}+\sqrt{3} \hat{j}) \times 2 B \hat{j}=M B(\hat{i} \times \hat{j})=M B(\hat{k})
$$
$\therefore$ Magnitude of torque is $\left|\tau_2\right|=M B=012 \mathrm{~N}-\mathrm{m}$
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