360417
A short bar magnet placed with its axis at \(30^{\circ}\) with a uniform external magnetic field of 0.25 \(T\) experiences a torque of magnitude equal to \(4.5 \times {10^{ - 2}}\;J\), then the magnitude of magnetic moment of the magnet is
1 \(0.86\;J/T\)
2 \(8.6\;J/T\)
3 \(0.36\;J/T\)
4 \(3.6\;J/T\)
Explanation:
Given, uniform magnetic field \(B = 0.25T\) The magnitude of torque is \(\tau=4.5 \times 10^{-2} J\) Angle between magnetic moment and magnetic field is \(\theta=30^{\circ}\) Torque experienced on a magnet placed in external magnetic field is \(\begin{aligned}& \tau=M \times B \\& \tau=M B \sin \theta \\& 4.5 \times 10^{-2}=M \times 0.25 \times \sin 30^{\circ} \\& M=\dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1}\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&= \dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1} \quad\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&=0.36 \mathrm{~J} / T\end{aligned}\)
NCERT Exemplar
PHXII05:MAGNETISM and MATTER
360418
A torque of \({10^{ - 5}}Nm\) is required to hold a magnet at with the horizontal component of the earth's magnetic field. The torque required to hold it at \(30^{\circ}\) will be:
1 \(\frac{1}{2} \times {10^{ - 5}}Nm\)
2 \(5 \times {10^{ - 6}}Nm\)
3 \(5\sqrt 3 \times {10^{ - 6}}Nm\)
4 Data is insufficient
Explanation:
The magnet in a magnetic field experiences a torque which rotates to a position in which the axis of the magnet is parallel to the field. \(\tau=M B \sin \theta\) Where, \(M\) is magnetic dipole moment, \(B\) the magnetic field and \(\theta\) the angle between the two. Given, \({\tau _1} = {10^{ - 5}}Nm,{\theta _1} = {90^ \circ }\) and \(\theta_{2}=30^{\circ}\) \({\tau _1} = MB\sin 90^\circ \,\,\,\,(1)\) \({\tau _2} = MB\sin 30^\circ \,\,\,(2)\) On dividing Eq. (1) by Eq. (2), we get \(\frac{{{\tau _1}}}{{{\tau _2}}} = \frac{{{{10}^{ - 5}}}}{{{\tau _2}}} = \frac{1}{{1/2}}\) \( \Rightarrow {\tau _2} = \frac{{{{10}^{ - 5}}}}{2} = \frac{{10}}{2} \times {10^{ - 6}} = 5 \times {10^{ - 6}}Nm\)
PHXII05:MAGNETISM and MATTER
360419
Assertion : Torque experienced by the bar magnet is maximum when field is applied perpendicular to magnetic moment. Reason : Torque on a bar magnet depends on the angle between applied magnetic field and magnetic dipole moment.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The torque \(\tau=M B \sin \theta\) [experienced by a bar magnet in a magnetic field \((B)\) with \(M\) as the magnetic dipole moment. Here, \(\theta\) is the angle between the magnetic moment and the magnetic field]. When \(\theta=90^{\circ}\) (perpendicular alignment), the torque is maximum \(\left(\because(\sin \theta)_{\max }=1\right)\). So correct option is (1).
PHXII05:MAGNETISM and MATTER
360420
A magnet of magnetic moment \(50 \hat{i} A-m^{2}\) is placed along the \(x\)-axis in a magnetic field \(\vec{B}=(0.5 \hat{i}+3.0 \hat{j}) T\). The torque acting on the magnet is
1 \(25 \sqrt{37} \hat{k} N-m\)
2 \(150 \hat{k} N-m\)
3 \(175 \hat{k} N-m\)
4 \(75 \hat{k} N-m\)
Explanation:
Torque acting on dipole \(\vec \tau = \vec M \times \vec B \Rightarrow \vec \tau = 50\hat i \times (0.5\hat i + 3\hat j)\) \( = 150(\hat i \times \hat j) = 150\hat k\,N - m\)
PHXII05:MAGNETISM and MATTER
360421
A magnetised needle of magnetic moment \(4.8 \times {10^{ - 2}}A{m^2}\) is placed at \(30^{\circ}\) with the direction of uniform magnetic field of magnitude \(3 \times {10^{ - 2}}\;T\). What is the torque acting on the needle?
360417
A short bar magnet placed with its axis at \(30^{\circ}\) with a uniform external magnetic field of 0.25 \(T\) experiences a torque of magnitude equal to \(4.5 \times {10^{ - 2}}\;J\), then the magnitude of magnetic moment of the magnet is
1 \(0.86\;J/T\)
2 \(8.6\;J/T\)
3 \(0.36\;J/T\)
4 \(3.6\;J/T\)
Explanation:
Given, uniform magnetic field \(B = 0.25T\) The magnitude of torque is \(\tau=4.5 \times 10^{-2} J\) Angle between magnetic moment and magnetic field is \(\theta=30^{\circ}\) Torque experienced on a magnet placed in external magnetic field is \(\begin{aligned}& \tau=M \times B \\& \tau=M B \sin \theta \\& 4.5 \times 10^{-2}=M \times 0.25 \times \sin 30^{\circ} \\& M=\dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1}\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&= \dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1} \quad\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&=0.36 \mathrm{~J} / T\end{aligned}\)
NCERT Exemplar
PHXII05:MAGNETISM and MATTER
360418
A torque of \({10^{ - 5}}Nm\) is required to hold a magnet at with the horizontal component of the earth's magnetic field. The torque required to hold it at \(30^{\circ}\) will be:
1 \(\frac{1}{2} \times {10^{ - 5}}Nm\)
2 \(5 \times {10^{ - 6}}Nm\)
3 \(5\sqrt 3 \times {10^{ - 6}}Nm\)
4 Data is insufficient
Explanation:
The magnet in a magnetic field experiences a torque which rotates to a position in which the axis of the magnet is parallel to the field. \(\tau=M B \sin \theta\) Where, \(M\) is magnetic dipole moment, \(B\) the magnetic field and \(\theta\) the angle between the two. Given, \({\tau _1} = {10^{ - 5}}Nm,{\theta _1} = {90^ \circ }\) and \(\theta_{2}=30^{\circ}\) \({\tau _1} = MB\sin 90^\circ \,\,\,\,(1)\) \({\tau _2} = MB\sin 30^\circ \,\,\,(2)\) On dividing Eq. (1) by Eq. (2), we get \(\frac{{{\tau _1}}}{{{\tau _2}}} = \frac{{{{10}^{ - 5}}}}{{{\tau _2}}} = \frac{1}{{1/2}}\) \( \Rightarrow {\tau _2} = \frac{{{{10}^{ - 5}}}}{2} = \frac{{10}}{2} \times {10^{ - 6}} = 5 \times {10^{ - 6}}Nm\)
PHXII05:MAGNETISM and MATTER
360419
Assertion : Torque experienced by the bar magnet is maximum when field is applied perpendicular to magnetic moment. Reason : Torque on a bar magnet depends on the angle between applied magnetic field and magnetic dipole moment.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The torque \(\tau=M B \sin \theta\) [experienced by a bar magnet in a magnetic field \((B)\) with \(M\) as the magnetic dipole moment. Here, \(\theta\) is the angle between the magnetic moment and the magnetic field]. When \(\theta=90^{\circ}\) (perpendicular alignment), the torque is maximum \(\left(\because(\sin \theta)_{\max }=1\right)\). So correct option is (1).
PHXII05:MAGNETISM and MATTER
360420
A magnet of magnetic moment \(50 \hat{i} A-m^{2}\) is placed along the \(x\)-axis in a magnetic field \(\vec{B}=(0.5 \hat{i}+3.0 \hat{j}) T\). The torque acting on the magnet is
1 \(25 \sqrt{37} \hat{k} N-m\)
2 \(150 \hat{k} N-m\)
3 \(175 \hat{k} N-m\)
4 \(75 \hat{k} N-m\)
Explanation:
Torque acting on dipole \(\vec \tau = \vec M \times \vec B \Rightarrow \vec \tau = 50\hat i \times (0.5\hat i + 3\hat j)\) \( = 150(\hat i \times \hat j) = 150\hat k\,N - m\)
PHXII05:MAGNETISM and MATTER
360421
A magnetised needle of magnetic moment \(4.8 \times {10^{ - 2}}A{m^2}\) is placed at \(30^{\circ}\) with the direction of uniform magnetic field of magnitude \(3 \times {10^{ - 2}}\;T\). What is the torque acting on the needle?
360417
A short bar magnet placed with its axis at \(30^{\circ}\) with a uniform external magnetic field of 0.25 \(T\) experiences a torque of magnitude equal to \(4.5 \times {10^{ - 2}}\;J\), then the magnitude of magnetic moment of the magnet is
1 \(0.86\;J/T\)
2 \(8.6\;J/T\)
3 \(0.36\;J/T\)
4 \(3.6\;J/T\)
Explanation:
Given, uniform magnetic field \(B = 0.25T\) The magnitude of torque is \(\tau=4.5 \times 10^{-2} J\) Angle between magnetic moment and magnetic field is \(\theta=30^{\circ}\) Torque experienced on a magnet placed in external magnetic field is \(\begin{aligned}& \tau=M \times B \\& \tau=M B \sin \theta \\& 4.5 \times 10^{-2}=M \times 0.25 \times \sin 30^{\circ} \\& M=\dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1}\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&= \dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1} \quad\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&=0.36 \mathrm{~J} / T\end{aligned}\)
NCERT Exemplar
PHXII05:MAGNETISM and MATTER
360418
A torque of \({10^{ - 5}}Nm\) is required to hold a magnet at with the horizontal component of the earth's magnetic field. The torque required to hold it at \(30^{\circ}\) will be:
1 \(\frac{1}{2} \times {10^{ - 5}}Nm\)
2 \(5 \times {10^{ - 6}}Nm\)
3 \(5\sqrt 3 \times {10^{ - 6}}Nm\)
4 Data is insufficient
Explanation:
The magnet in a magnetic field experiences a torque which rotates to a position in which the axis of the magnet is parallel to the field. \(\tau=M B \sin \theta\) Where, \(M\) is magnetic dipole moment, \(B\) the magnetic field and \(\theta\) the angle between the two. Given, \({\tau _1} = {10^{ - 5}}Nm,{\theta _1} = {90^ \circ }\) and \(\theta_{2}=30^{\circ}\) \({\tau _1} = MB\sin 90^\circ \,\,\,\,(1)\) \({\tau _2} = MB\sin 30^\circ \,\,\,(2)\) On dividing Eq. (1) by Eq. (2), we get \(\frac{{{\tau _1}}}{{{\tau _2}}} = \frac{{{{10}^{ - 5}}}}{{{\tau _2}}} = \frac{1}{{1/2}}\) \( \Rightarrow {\tau _2} = \frac{{{{10}^{ - 5}}}}{2} = \frac{{10}}{2} \times {10^{ - 6}} = 5 \times {10^{ - 6}}Nm\)
PHXII05:MAGNETISM and MATTER
360419
Assertion : Torque experienced by the bar magnet is maximum when field is applied perpendicular to magnetic moment. Reason : Torque on a bar magnet depends on the angle between applied magnetic field and magnetic dipole moment.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The torque \(\tau=M B \sin \theta\) [experienced by a bar magnet in a magnetic field \((B)\) with \(M\) as the magnetic dipole moment. Here, \(\theta\) is the angle between the magnetic moment and the magnetic field]. When \(\theta=90^{\circ}\) (perpendicular alignment), the torque is maximum \(\left(\because(\sin \theta)_{\max }=1\right)\). So correct option is (1).
PHXII05:MAGNETISM and MATTER
360420
A magnet of magnetic moment \(50 \hat{i} A-m^{2}\) is placed along the \(x\)-axis in a magnetic field \(\vec{B}=(0.5 \hat{i}+3.0 \hat{j}) T\). The torque acting on the magnet is
1 \(25 \sqrt{37} \hat{k} N-m\)
2 \(150 \hat{k} N-m\)
3 \(175 \hat{k} N-m\)
4 \(75 \hat{k} N-m\)
Explanation:
Torque acting on dipole \(\vec \tau = \vec M \times \vec B \Rightarrow \vec \tau = 50\hat i \times (0.5\hat i + 3\hat j)\) \( = 150(\hat i \times \hat j) = 150\hat k\,N - m\)
PHXII05:MAGNETISM and MATTER
360421
A magnetised needle of magnetic moment \(4.8 \times {10^{ - 2}}A{m^2}\) is placed at \(30^{\circ}\) with the direction of uniform magnetic field of magnitude \(3 \times {10^{ - 2}}\;T\). What is the torque acting on the needle?
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PHXII05:MAGNETISM and MATTER
360417
A short bar magnet placed with its axis at \(30^{\circ}\) with a uniform external magnetic field of 0.25 \(T\) experiences a torque of magnitude equal to \(4.5 \times {10^{ - 2}}\;J\), then the magnitude of magnetic moment of the magnet is
1 \(0.86\;J/T\)
2 \(8.6\;J/T\)
3 \(0.36\;J/T\)
4 \(3.6\;J/T\)
Explanation:
Given, uniform magnetic field \(B = 0.25T\) The magnitude of torque is \(\tau=4.5 \times 10^{-2} J\) Angle between magnetic moment and magnetic field is \(\theta=30^{\circ}\) Torque experienced on a magnet placed in external magnetic field is \(\begin{aligned}& \tau=M \times B \\& \tau=M B \sin \theta \\& 4.5 \times 10^{-2}=M \times 0.25 \times \sin 30^{\circ} \\& M=\dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1}\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&= \dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1} \quad\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&=0.36 \mathrm{~J} / T\end{aligned}\)
NCERT Exemplar
PHXII05:MAGNETISM and MATTER
360418
A torque of \({10^{ - 5}}Nm\) is required to hold a magnet at with the horizontal component of the earth's magnetic field. The torque required to hold it at \(30^{\circ}\) will be:
1 \(\frac{1}{2} \times {10^{ - 5}}Nm\)
2 \(5 \times {10^{ - 6}}Nm\)
3 \(5\sqrt 3 \times {10^{ - 6}}Nm\)
4 Data is insufficient
Explanation:
The magnet in a magnetic field experiences a torque which rotates to a position in which the axis of the magnet is parallel to the field. \(\tau=M B \sin \theta\) Where, \(M\) is magnetic dipole moment, \(B\) the magnetic field and \(\theta\) the angle between the two. Given, \({\tau _1} = {10^{ - 5}}Nm,{\theta _1} = {90^ \circ }\) and \(\theta_{2}=30^{\circ}\) \({\tau _1} = MB\sin 90^\circ \,\,\,\,(1)\) \({\tau _2} = MB\sin 30^\circ \,\,\,(2)\) On dividing Eq. (1) by Eq. (2), we get \(\frac{{{\tau _1}}}{{{\tau _2}}} = \frac{{{{10}^{ - 5}}}}{{{\tau _2}}} = \frac{1}{{1/2}}\) \( \Rightarrow {\tau _2} = \frac{{{{10}^{ - 5}}}}{2} = \frac{{10}}{2} \times {10^{ - 6}} = 5 \times {10^{ - 6}}Nm\)
PHXII05:MAGNETISM and MATTER
360419
Assertion : Torque experienced by the bar magnet is maximum when field is applied perpendicular to magnetic moment. Reason : Torque on a bar magnet depends on the angle between applied magnetic field and magnetic dipole moment.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The torque \(\tau=M B \sin \theta\) [experienced by a bar magnet in a magnetic field \((B)\) with \(M\) as the magnetic dipole moment. Here, \(\theta\) is the angle between the magnetic moment and the magnetic field]. When \(\theta=90^{\circ}\) (perpendicular alignment), the torque is maximum \(\left(\because(\sin \theta)_{\max }=1\right)\). So correct option is (1).
PHXII05:MAGNETISM and MATTER
360420
A magnet of magnetic moment \(50 \hat{i} A-m^{2}\) is placed along the \(x\)-axis in a magnetic field \(\vec{B}=(0.5 \hat{i}+3.0 \hat{j}) T\). The torque acting on the magnet is
1 \(25 \sqrt{37} \hat{k} N-m\)
2 \(150 \hat{k} N-m\)
3 \(175 \hat{k} N-m\)
4 \(75 \hat{k} N-m\)
Explanation:
Torque acting on dipole \(\vec \tau = \vec M \times \vec B \Rightarrow \vec \tau = 50\hat i \times (0.5\hat i + 3\hat j)\) \( = 150(\hat i \times \hat j) = 150\hat k\,N - m\)
PHXII05:MAGNETISM and MATTER
360421
A magnetised needle of magnetic moment \(4.8 \times {10^{ - 2}}A{m^2}\) is placed at \(30^{\circ}\) with the direction of uniform magnetic field of magnitude \(3 \times {10^{ - 2}}\;T\). What is the torque acting on the needle?
360417
A short bar magnet placed with its axis at \(30^{\circ}\) with a uniform external magnetic field of 0.25 \(T\) experiences a torque of magnitude equal to \(4.5 \times {10^{ - 2}}\;J\), then the magnitude of magnetic moment of the magnet is
1 \(0.86\;J/T\)
2 \(8.6\;J/T\)
3 \(0.36\;J/T\)
4 \(3.6\;J/T\)
Explanation:
Given, uniform magnetic field \(B = 0.25T\) The magnitude of torque is \(\tau=4.5 \times 10^{-2} J\) Angle between magnetic moment and magnetic field is \(\theta=30^{\circ}\) Torque experienced on a magnet placed in external magnetic field is \(\begin{aligned}& \tau=M \times B \\& \tau=M B \sin \theta \\& 4.5 \times 10^{-2}=M \times 0.25 \times \sin 30^{\circ} \\& M=\dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1}\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&= \dfrac{4.5 \times 10^{-2} \times 2}{0.25 \times 1} \quad\left(\because \sin 30^{\circ}=\dfrac{1}{2}\right) \\&=0.36 \mathrm{~J} / T\end{aligned}\)
NCERT Exemplar
PHXII05:MAGNETISM and MATTER
360418
A torque of \({10^{ - 5}}Nm\) is required to hold a magnet at with the horizontal component of the earth's magnetic field. The torque required to hold it at \(30^{\circ}\) will be:
1 \(\frac{1}{2} \times {10^{ - 5}}Nm\)
2 \(5 \times {10^{ - 6}}Nm\)
3 \(5\sqrt 3 \times {10^{ - 6}}Nm\)
4 Data is insufficient
Explanation:
The magnet in a magnetic field experiences a torque which rotates to a position in which the axis of the magnet is parallel to the field. \(\tau=M B \sin \theta\) Where, \(M\) is magnetic dipole moment, \(B\) the magnetic field and \(\theta\) the angle between the two. Given, \({\tau _1} = {10^{ - 5}}Nm,{\theta _1} = {90^ \circ }\) and \(\theta_{2}=30^{\circ}\) \({\tau _1} = MB\sin 90^\circ \,\,\,\,(1)\) \({\tau _2} = MB\sin 30^\circ \,\,\,(2)\) On dividing Eq. (1) by Eq. (2), we get \(\frac{{{\tau _1}}}{{{\tau _2}}} = \frac{{{{10}^{ - 5}}}}{{{\tau _2}}} = \frac{1}{{1/2}}\) \( \Rightarrow {\tau _2} = \frac{{{{10}^{ - 5}}}}{2} = \frac{{10}}{2} \times {10^{ - 6}} = 5 \times {10^{ - 6}}Nm\)
PHXII05:MAGNETISM and MATTER
360419
Assertion : Torque experienced by the bar magnet is maximum when field is applied perpendicular to magnetic moment. Reason : Torque on a bar magnet depends on the angle between applied magnetic field and magnetic dipole moment.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The torque \(\tau=M B \sin \theta\) [experienced by a bar magnet in a magnetic field \((B)\) with \(M\) as the magnetic dipole moment. Here, \(\theta\) is the angle between the magnetic moment and the magnetic field]. When \(\theta=90^{\circ}\) (perpendicular alignment), the torque is maximum \(\left(\because(\sin \theta)_{\max }=1\right)\). So correct option is (1).
PHXII05:MAGNETISM and MATTER
360420
A magnet of magnetic moment \(50 \hat{i} A-m^{2}\) is placed along the \(x\)-axis in a magnetic field \(\vec{B}=(0.5 \hat{i}+3.0 \hat{j}) T\). The torque acting on the magnet is
1 \(25 \sqrt{37} \hat{k} N-m\)
2 \(150 \hat{k} N-m\)
3 \(175 \hat{k} N-m\)
4 \(75 \hat{k} N-m\)
Explanation:
Torque acting on dipole \(\vec \tau = \vec M \times \vec B \Rightarrow \vec \tau = 50\hat i \times (0.5\hat i + 3\hat j)\) \( = 150(\hat i \times \hat j) = 150\hat k\,N - m\)
PHXII05:MAGNETISM and MATTER
360421
A magnetised needle of magnetic moment \(4.8 \times {10^{ - 2}}A{m^2}\) is placed at \(30^{\circ}\) with the direction of uniform magnetic field of magnitude \(3 \times {10^{ - 2}}\;T\). What is the torque acting on the needle?
