153744
A short bar magnet is placed with its axis at $30^{\circ}$ with an external field $0.05 \mathrm{~T}$. If the magnetic moment of the magnet is $0.8 \mathrm{Am}^{2}$, then the torque experienced by the magnet is
1 $0.10 \mathrm{~N} . \mathrm{m}$
2 $0.08 \mathrm{~N} . \mathrm{m}$
3 0.01 N.m
4 $0.02 \mathrm{~N} . \mathrm{m}$
Explanation:
D Given, $\theta=30^{\circ}, \quad \mathrm{B}=0.05 \mathrm{~T}, \quad \mathrm{M}=0.8 \mathrm{Am}^{2}$ Torque experienced by a bar magnet placed at an angle $\theta$ $\tau =\mathrm{MB} \sin \theta$ $=0.8 \times 0.05 \times \sin 30^{\circ}$ $=0.8 \times 0.05 \times \frac{1}{2}$ $\tau =0.02 \mathrm{Nm}$
TS EAMCET 31.07.2022
Moving Charges & Magnetism
153747
What is the net force on the square coil?
1 $25 \times 10^{-7} \mathrm{~N}$ moving towards wire
2 $25 \times 10^{-7} \mathrm{~N}$ moving away from wire
3 $35 \times 10^{-7} \mathrm{~N}$ moving towards wire
4 $35 \times 10^{-7} \mathrm{~N}$ moving away from wire
153748
A 100 turns coil shown in figure carries a current of $2 \mathrm{~A}$ in a magnetic field $\mathrm{B}=0.2 \mathrm{~Wb}$. $\mathrm{m}^{-2}$ The torque acting on the coil is
1 $0.32 \mathrm{~N}$. m tending to rotate the side $\mathrm{AD}$ out of the page
2 $0.32 \mathrm{~N} . \mathrm{m}$ tending to rotate the side $\mathrm{AD}$ into the page
3 $0.0032 \mathrm{~N}$. $\mathrm{m}$ tending to rotate the side $\mathrm{AD}$ out of the page
4 $0.0032 \mathrm{~N} . \mathrm{m}$ tending to rotate the side $\mathrm{AD}$ into the page
Explanation:
A Given, $\mathrm{N}=100 \text { turns; } \mathrm{I}=2 \mathrm{~A}, \quad \mathrm{~B}=0.2 \mathrm{~Wb} \mathrm{~m}^{-2}$ The torque experienced by the coil is, $\tau =\text { NBIA } \quad\left(\because \mathrm{A}=0.08 \times 0.1 \mathrm{~m}^{2}\right)$ $\tau =100 \times 0.2 \times 2 \times(0.08 \times 0.1)$ $\tau =0.32 \mathrm{Nm}$ Direction can be found by Fleming's Left Hand Rule. Hence, $0.32 \mathrm{~N}-\mathrm{m}$ tending to rotate the side AD out of the page.
AP EAMCET-03.09.2021
Moving Charges & Magnetism
153749
The rectangular coil of area $A$ is in a field $B$. Find the torque about the $Z$-axis when the coil lies in the position shown and carries a current $I$.
1 $\mathrm{IAB}$ in negative $Z$-axis
2 IAB in positive $Z$-axis
3 $2 \mathrm{IAB}$ in positive $Z$-axis
4 2IAB in negative $Z$-axis
Explanation:
B Torque, $\tau =\mathrm{I}(\mathrm{A} \hat{\mathrm{i}} \times \mathrm{B} \hat{j})$ $\tau =\mathrm{IAB} \hat{k}$ Hence, $\tau$ is IAB in positive $Z$-axis
153744
A short bar magnet is placed with its axis at $30^{\circ}$ with an external field $0.05 \mathrm{~T}$. If the magnetic moment of the magnet is $0.8 \mathrm{Am}^{2}$, then the torque experienced by the magnet is
1 $0.10 \mathrm{~N} . \mathrm{m}$
2 $0.08 \mathrm{~N} . \mathrm{m}$
3 0.01 N.m
4 $0.02 \mathrm{~N} . \mathrm{m}$
Explanation:
D Given, $\theta=30^{\circ}, \quad \mathrm{B}=0.05 \mathrm{~T}, \quad \mathrm{M}=0.8 \mathrm{Am}^{2}$ Torque experienced by a bar magnet placed at an angle $\theta$ $\tau =\mathrm{MB} \sin \theta$ $=0.8 \times 0.05 \times \sin 30^{\circ}$ $=0.8 \times 0.05 \times \frac{1}{2}$ $\tau =0.02 \mathrm{Nm}$
TS EAMCET 31.07.2022
Moving Charges & Magnetism
153747
What is the net force on the square coil?
1 $25 \times 10^{-7} \mathrm{~N}$ moving towards wire
2 $25 \times 10^{-7} \mathrm{~N}$ moving away from wire
3 $35 \times 10^{-7} \mathrm{~N}$ moving towards wire
4 $35 \times 10^{-7} \mathrm{~N}$ moving away from wire
153748
A 100 turns coil shown in figure carries a current of $2 \mathrm{~A}$ in a magnetic field $\mathrm{B}=0.2 \mathrm{~Wb}$. $\mathrm{m}^{-2}$ The torque acting on the coil is
1 $0.32 \mathrm{~N}$. m tending to rotate the side $\mathrm{AD}$ out of the page
2 $0.32 \mathrm{~N} . \mathrm{m}$ tending to rotate the side $\mathrm{AD}$ into the page
3 $0.0032 \mathrm{~N}$. $\mathrm{m}$ tending to rotate the side $\mathrm{AD}$ out of the page
4 $0.0032 \mathrm{~N} . \mathrm{m}$ tending to rotate the side $\mathrm{AD}$ into the page
Explanation:
A Given, $\mathrm{N}=100 \text { turns; } \mathrm{I}=2 \mathrm{~A}, \quad \mathrm{~B}=0.2 \mathrm{~Wb} \mathrm{~m}^{-2}$ The torque experienced by the coil is, $\tau =\text { NBIA } \quad\left(\because \mathrm{A}=0.08 \times 0.1 \mathrm{~m}^{2}\right)$ $\tau =100 \times 0.2 \times 2 \times(0.08 \times 0.1)$ $\tau =0.32 \mathrm{Nm}$ Direction can be found by Fleming's Left Hand Rule. Hence, $0.32 \mathrm{~N}-\mathrm{m}$ tending to rotate the side AD out of the page.
AP EAMCET-03.09.2021
Moving Charges & Magnetism
153749
The rectangular coil of area $A$ is in a field $B$. Find the torque about the $Z$-axis when the coil lies in the position shown and carries a current $I$.
1 $\mathrm{IAB}$ in negative $Z$-axis
2 IAB in positive $Z$-axis
3 $2 \mathrm{IAB}$ in positive $Z$-axis
4 2IAB in negative $Z$-axis
Explanation:
B Torque, $\tau =\mathrm{I}(\mathrm{A} \hat{\mathrm{i}} \times \mathrm{B} \hat{j})$ $\tau =\mathrm{IAB} \hat{k}$ Hence, $\tau$ is IAB in positive $Z$-axis
153744
A short bar magnet is placed with its axis at $30^{\circ}$ with an external field $0.05 \mathrm{~T}$. If the magnetic moment of the magnet is $0.8 \mathrm{Am}^{2}$, then the torque experienced by the magnet is
1 $0.10 \mathrm{~N} . \mathrm{m}$
2 $0.08 \mathrm{~N} . \mathrm{m}$
3 0.01 N.m
4 $0.02 \mathrm{~N} . \mathrm{m}$
Explanation:
D Given, $\theta=30^{\circ}, \quad \mathrm{B}=0.05 \mathrm{~T}, \quad \mathrm{M}=0.8 \mathrm{Am}^{2}$ Torque experienced by a bar magnet placed at an angle $\theta$ $\tau =\mathrm{MB} \sin \theta$ $=0.8 \times 0.05 \times \sin 30^{\circ}$ $=0.8 \times 0.05 \times \frac{1}{2}$ $\tau =0.02 \mathrm{Nm}$
TS EAMCET 31.07.2022
Moving Charges & Magnetism
153747
What is the net force on the square coil?
1 $25 \times 10^{-7} \mathrm{~N}$ moving towards wire
2 $25 \times 10^{-7} \mathrm{~N}$ moving away from wire
3 $35 \times 10^{-7} \mathrm{~N}$ moving towards wire
4 $35 \times 10^{-7} \mathrm{~N}$ moving away from wire
153748
A 100 turns coil shown in figure carries a current of $2 \mathrm{~A}$ in a magnetic field $\mathrm{B}=0.2 \mathrm{~Wb}$. $\mathrm{m}^{-2}$ The torque acting on the coil is
1 $0.32 \mathrm{~N}$. m tending to rotate the side $\mathrm{AD}$ out of the page
2 $0.32 \mathrm{~N} . \mathrm{m}$ tending to rotate the side $\mathrm{AD}$ into the page
3 $0.0032 \mathrm{~N}$. $\mathrm{m}$ tending to rotate the side $\mathrm{AD}$ out of the page
4 $0.0032 \mathrm{~N} . \mathrm{m}$ tending to rotate the side $\mathrm{AD}$ into the page
Explanation:
A Given, $\mathrm{N}=100 \text { turns; } \mathrm{I}=2 \mathrm{~A}, \quad \mathrm{~B}=0.2 \mathrm{~Wb} \mathrm{~m}^{-2}$ The torque experienced by the coil is, $\tau =\text { NBIA } \quad\left(\because \mathrm{A}=0.08 \times 0.1 \mathrm{~m}^{2}\right)$ $\tau =100 \times 0.2 \times 2 \times(0.08 \times 0.1)$ $\tau =0.32 \mathrm{Nm}$ Direction can be found by Fleming's Left Hand Rule. Hence, $0.32 \mathrm{~N}-\mathrm{m}$ tending to rotate the side AD out of the page.
AP EAMCET-03.09.2021
Moving Charges & Magnetism
153749
The rectangular coil of area $A$ is in a field $B$. Find the torque about the $Z$-axis when the coil lies in the position shown and carries a current $I$.
1 $\mathrm{IAB}$ in negative $Z$-axis
2 IAB in positive $Z$-axis
3 $2 \mathrm{IAB}$ in positive $Z$-axis
4 2IAB in negative $Z$-axis
Explanation:
B Torque, $\tau =\mathrm{I}(\mathrm{A} \hat{\mathrm{i}} \times \mathrm{B} \hat{j})$ $\tau =\mathrm{IAB} \hat{k}$ Hence, $\tau$ is IAB in positive $Z$-axis
153744
A short bar magnet is placed with its axis at $30^{\circ}$ with an external field $0.05 \mathrm{~T}$. If the magnetic moment of the magnet is $0.8 \mathrm{Am}^{2}$, then the torque experienced by the magnet is
1 $0.10 \mathrm{~N} . \mathrm{m}$
2 $0.08 \mathrm{~N} . \mathrm{m}$
3 0.01 N.m
4 $0.02 \mathrm{~N} . \mathrm{m}$
Explanation:
D Given, $\theta=30^{\circ}, \quad \mathrm{B}=0.05 \mathrm{~T}, \quad \mathrm{M}=0.8 \mathrm{Am}^{2}$ Torque experienced by a bar magnet placed at an angle $\theta$ $\tau =\mathrm{MB} \sin \theta$ $=0.8 \times 0.05 \times \sin 30^{\circ}$ $=0.8 \times 0.05 \times \frac{1}{2}$ $\tau =0.02 \mathrm{Nm}$
TS EAMCET 31.07.2022
Moving Charges & Magnetism
153747
What is the net force on the square coil?
1 $25 \times 10^{-7} \mathrm{~N}$ moving towards wire
2 $25 \times 10^{-7} \mathrm{~N}$ moving away from wire
3 $35 \times 10^{-7} \mathrm{~N}$ moving towards wire
4 $35 \times 10^{-7} \mathrm{~N}$ moving away from wire
153748
A 100 turns coil shown in figure carries a current of $2 \mathrm{~A}$ in a magnetic field $\mathrm{B}=0.2 \mathrm{~Wb}$. $\mathrm{m}^{-2}$ The torque acting on the coil is
1 $0.32 \mathrm{~N}$. m tending to rotate the side $\mathrm{AD}$ out of the page
2 $0.32 \mathrm{~N} . \mathrm{m}$ tending to rotate the side $\mathrm{AD}$ into the page
3 $0.0032 \mathrm{~N}$. $\mathrm{m}$ tending to rotate the side $\mathrm{AD}$ out of the page
4 $0.0032 \mathrm{~N} . \mathrm{m}$ tending to rotate the side $\mathrm{AD}$ into the page
Explanation:
A Given, $\mathrm{N}=100 \text { turns; } \mathrm{I}=2 \mathrm{~A}, \quad \mathrm{~B}=0.2 \mathrm{~Wb} \mathrm{~m}^{-2}$ The torque experienced by the coil is, $\tau =\text { NBIA } \quad\left(\because \mathrm{A}=0.08 \times 0.1 \mathrm{~m}^{2}\right)$ $\tau =100 \times 0.2 \times 2 \times(0.08 \times 0.1)$ $\tau =0.32 \mathrm{Nm}$ Direction can be found by Fleming's Left Hand Rule. Hence, $0.32 \mathrm{~N}-\mathrm{m}$ tending to rotate the side AD out of the page.
AP EAMCET-03.09.2021
Moving Charges & Magnetism
153749
The rectangular coil of area $A$ is in a field $B$. Find the torque about the $Z$-axis when the coil lies in the position shown and carries a current $I$.
1 $\mathrm{IAB}$ in negative $Z$-axis
2 IAB in positive $Z$-axis
3 $2 \mathrm{IAB}$ in positive $Z$-axis
4 2IAB in negative $Z$-axis
Explanation:
B Torque, $\tau =\mathrm{I}(\mathrm{A} \hat{\mathrm{i}} \times \mathrm{B} \hat{j})$ $\tau =\mathrm{IAB} \hat{k}$ Hence, $\tau$ is IAB in positive $Z$-axis
