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A short bar magnet, placed with its axis at \({30^{\circ}}\) with a uniform external magnetic field of \({0.16 T}\) experiences a torque of magnitude 0.032 J . The magnetic moment of the bar magnet will be
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A bar magnet of magnetic moment \(200\;A - {m^2}\) is suspended in a magnetic field of intensity \(0.25\;N/A - m\). The couple required to deflect it through \(30^{\circ}\) is
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A bar magnet of moment \(M=\hat{i}+\hat{j}\) is placed in a magnetic field induction \(\vec{B}=3 \hat{i}+4 \hat{j}+4 \hat{k}\). The torque acting on the magnet is
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A magnetic needle lying parallel to a magnetic field required \(W\) units of work to turn it through \(90^{\circ}\). The torque required to maintain the needle in this position will be
1 \(W\)
2 2\(W\)
3 \(\sqrt{3} W\)
4 \(\dfrac{\sqrt{3}}{2} W\)
Explanation:
Work done \(W=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)\) \(=M B\left(\cos 0^{\circ}-\cos 90^{\circ}\right)\) \(\begin{aligned}& =M B(1-0)=M B \\& \text { and } \tau=M \cdot B \sin \theta=M B \sin 90^{\circ}=M B \\& \tau=W\end{aligned}\)
360426
A short bar magnet, placed with its axis at \({30^{\circ}}\) with a uniform external magnetic field of \({0.16 T}\) experiences a torque of magnitude 0.032 J . The magnetic moment of the bar magnet will be
360427
A bar magnet of magnetic moment \(200\;A - {m^2}\) is suspended in a magnetic field of intensity \(0.25\;N/A - m\). The couple required to deflect it through \(30^{\circ}\) is
360428
A bar magnet of moment \(M=\hat{i}+\hat{j}\) is placed in a magnetic field induction \(\vec{B}=3 \hat{i}+4 \hat{j}+4 \hat{k}\). The torque acting on the magnet is
360429
A magnetic needle lying parallel to a magnetic field required \(W\) units of work to turn it through \(90^{\circ}\). The torque required to maintain the needle in this position will be
1 \(W\)
2 2\(W\)
3 \(\sqrt{3} W\)
4 \(\dfrac{\sqrt{3}}{2} W\)
Explanation:
Work done \(W=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)\) \(=M B\left(\cos 0^{\circ}-\cos 90^{\circ}\right)\) \(\begin{aligned}& =M B(1-0)=M B \\& \text { and } \tau=M \cdot B \sin \theta=M B \sin 90^{\circ}=M B \\& \tau=W\end{aligned}\)
360426
A short bar magnet, placed with its axis at \({30^{\circ}}\) with a uniform external magnetic field of \({0.16 T}\) experiences a torque of magnitude 0.032 J . The magnetic moment of the bar magnet will be
360427
A bar magnet of magnetic moment \(200\;A - {m^2}\) is suspended in a magnetic field of intensity \(0.25\;N/A - m\). The couple required to deflect it through \(30^{\circ}\) is
360428
A bar magnet of moment \(M=\hat{i}+\hat{j}\) is placed in a magnetic field induction \(\vec{B}=3 \hat{i}+4 \hat{j}+4 \hat{k}\). The torque acting on the magnet is
360429
A magnetic needle lying parallel to a magnetic field required \(W\) units of work to turn it through \(90^{\circ}\). The torque required to maintain the needle in this position will be
1 \(W\)
2 2\(W\)
3 \(\sqrt{3} W\)
4 \(\dfrac{\sqrt{3}}{2} W\)
Explanation:
Work done \(W=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)\) \(=M B\left(\cos 0^{\circ}-\cos 90^{\circ}\right)\) \(\begin{aligned}& =M B(1-0)=M B \\& \text { and } \tau=M \cdot B \sin \theta=M B \sin 90^{\circ}=M B \\& \tau=W\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII05:MAGNETISM and MATTER
360426
A short bar magnet, placed with its axis at \({30^{\circ}}\) with a uniform external magnetic field of \({0.16 T}\) experiences a torque of magnitude 0.032 J . The magnetic moment of the bar magnet will be
360427
A bar magnet of magnetic moment \(200\;A - {m^2}\) is suspended in a magnetic field of intensity \(0.25\;N/A - m\). The couple required to deflect it through \(30^{\circ}\) is
360428
A bar magnet of moment \(M=\hat{i}+\hat{j}\) is placed in a magnetic field induction \(\vec{B}=3 \hat{i}+4 \hat{j}+4 \hat{k}\). The torque acting on the magnet is
360429
A magnetic needle lying parallel to a magnetic field required \(W\) units of work to turn it through \(90^{\circ}\). The torque required to maintain the needle in this position will be
1 \(W\)
2 2\(W\)
3 \(\sqrt{3} W\)
4 \(\dfrac{\sqrt{3}}{2} W\)
Explanation:
Work done \(W=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)\) \(=M B\left(\cos 0^{\circ}-\cos 90^{\circ}\right)\) \(\begin{aligned}& =M B(1-0)=M B \\& \text { and } \tau=M \cdot B \sin \theta=M B \sin 90^{\circ}=M B \\& \tau=W\end{aligned}\)