153820
For the magnetic field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be
1 $0^{\circ}$
2 $90^{\circ}$
3 $180^{\circ}$
4 $45^{\circ}$
Explanation:
B The magnetic field due to small element of conductor of length $\mathrm{d} l$ is given by $\mathrm{dB}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{Id} l \sin \theta}{\mathrm{r}^{2}}$ This value will be maximum when, $\sin \theta=1$ $\sin \theta=\sin 90^{\circ}$ $\therefore \quad \theta=90^{\circ}$ Moving Charges \& Magnetism
J and K CET- 2005
Moving Charges & Magnetism
153821
A bar magnet has a magnetic moment of 200 $\mathrm{Am}^{2}$. The magnet is suspended in a magnetic field of $0.30 \mathrm{NA}^{-1} \mathrm{~m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$, will be
153824
A magnet of moment $2 \mathrm{~A}-\mathrm{m}^{2}$ is placed in a uniform magnetic field of $5 \mathrm{~Wb} / \mathrm{m}^{2}$. If the magnet experiences a torque of $5 \mathrm{~N}$, then the angle between the direction of magnetic field and magnet is
1 $\frac{\pi}{6}$
2 $\frac{\pi}{4}$
3 $\frac{\pi}{3}$
4 $\frac{\pi}{2}$
Explanation:
A Given that , $\mathrm{M}=2 \mathrm{~A}-\mathrm{m}^{2}$ $\tau=5 \mathrm{~N} \quad \mathrm{~B}=5 \mathrm{~Wb} / \mathrm{m}^{2}$ We know that, Torque $\tau=\mathrm{MB} \sin \theta$ $5=2 \times 5 \sin \theta$ $5=10 \sin \theta$ $\sin \theta=\frac{1}{2}=\sin \frac{\pi}{6}$ $\theta=\pi / 6$
SRMJEEE - 2012
Moving Charges & Magnetism
153825
A current carrying straight wire is kept along the axis of a circular loop carrying a current. The straight wire
1 Will exert an inward force on the circular loop
2 Will exert an outward force on the circular loop
3 Will exert a force on the circular loop parallel to itself
4 Will not exert any force on the circular loop
Explanation:
D The magnetic force on a wire carrying an electric current $\mathrm{i}$ is given as $\overrightarrow{\mathrm{F}}=\mathrm{i}(\vec{l} \times \overrightarrow{\mathrm{B}})$ Where, $l=$ Length of the wire $\mathrm{B}=$ Magnetic field acting on wire If a current carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right hand thumb rule, the magnetic field due to wire on the current carrying loop will be along its circumference, which contains a current element $i \overrightarrow{\mathrm{d} l}$. So, the cross product will be $(\vec{l} \times \overrightarrow{\mathrm{B}})=0$ $\therefore \quad \overrightarrow{\mathrm{F}}=0$ So, the straight wire will not exert any force on the circular loop.
153820
For the magnetic field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be
1 $0^{\circ}$
2 $90^{\circ}$
3 $180^{\circ}$
4 $45^{\circ}$
Explanation:
B The magnetic field due to small element of conductor of length $\mathrm{d} l$ is given by $\mathrm{dB}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{Id} l \sin \theta}{\mathrm{r}^{2}}$ This value will be maximum when, $\sin \theta=1$ $\sin \theta=\sin 90^{\circ}$ $\therefore \quad \theta=90^{\circ}$ Moving Charges \& Magnetism
J and K CET- 2005
Moving Charges & Magnetism
153821
A bar magnet has a magnetic moment of 200 $\mathrm{Am}^{2}$. The magnet is suspended in a magnetic field of $0.30 \mathrm{NA}^{-1} \mathrm{~m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$, will be
153824
A magnet of moment $2 \mathrm{~A}-\mathrm{m}^{2}$ is placed in a uniform magnetic field of $5 \mathrm{~Wb} / \mathrm{m}^{2}$. If the magnet experiences a torque of $5 \mathrm{~N}$, then the angle between the direction of magnetic field and magnet is
1 $\frac{\pi}{6}$
2 $\frac{\pi}{4}$
3 $\frac{\pi}{3}$
4 $\frac{\pi}{2}$
Explanation:
A Given that , $\mathrm{M}=2 \mathrm{~A}-\mathrm{m}^{2}$ $\tau=5 \mathrm{~N} \quad \mathrm{~B}=5 \mathrm{~Wb} / \mathrm{m}^{2}$ We know that, Torque $\tau=\mathrm{MB} \sin \theta$ $5=2 \times 5 \sin \theta$ $5=10 \sin \theta$ $\sin \theta=\frac{1}{2}=\sin \frac{\pi}{6}$ $\theta=\pi / 6$
SRMJEEE - 2012
Moving Charges & Magnetism
153825
A current carrying straight wire is kept along the axis of a circular loop carrying a current. The straight wire
1 Will exert an inward force on the circular loop
2 Will exert an outward force on the circular loop
3 Will exert a force on the circular loop parallel to itself
4 Will not exert any force on the circular loop
Explanation:
D The magnetic force on a wire carrying an electric current $\mathrm{i}$ is given as $\overrightarrow{\mathrm{F}}=\mathrm{i}(\vec{l} \times \overrightarrow{\mathrm{B}})$ Where, $l=$ Length of the wire $\mathrm{B}=$ Magnetic field acting on wire If a current carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right hand thumb rule, the magnetic field due to wire on the current carrying loop will be along its circumference, which contains a current element $i \overrightarrow{\mathrm{d} l}$. So, the cross product will be $(\vec{l} \times \overrightarrow{\mathrm{B}})=0$ $\therefore \quad \overrightarrow{\mathrm{F}}=0$ So, the straight wire will not exert any force on the circular loop.
153820
For the magnetic field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be
1 $0^{\circ}$
2 $90^{\circ}$
3 $180^{\circ}$
4 $45^{\circ}$
Explanation:
B The magnetic field due to small element of conductor of length $\mathrm{d} l$ is given by $\mathrm{dB}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{Id} l \sin \theta}{\mathrm{r}^{2}}$ This value will be maximum when, $\sin \theta=1$ $\sin \theta=\sin 90^{\circ}$ $\therefore \quad \theta=90^{\circ}$ Moving Charges \& Magnetism
J and K CET- 2005
Moving Charges & Magnetism
153821
A bar magnet has a magnetic moment of 200 $\mathrm{Am}^{2}$. The magnet is suspended in a magnetic field of $0.30 \mathrm{NA}^{-1} \mathrm{~m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$, will be
153824
A magnet of moment $2 \mathrm{~A}-\mathrm{m}^{2}$ is placed in a uniform magnetic field of $5 \mathrm{~Wb} / \mathrm{m}^{2}$. If the magnet experiences a torque of $5 \mathrm{~N}$, then the angle between the direction of magnetic field and magnet is
1 $\frac{\pi}{6}$
2 $\frac{\pi}{4}$
3 $\frac{\pi}{3}$
4 $\frac{\pi}{2}$
Explanation:
A Given that , $\mathrm{M}=2 \mathrm{~A}-\mathrm{m}^{2}$ $\tau=5 \mathrm{~N} \quad \mathrm{~B}=5 \mathrm{~Wb} / \mathrm{m}^{2}$ We know that, Torque $\tau=\mathrm{MB} \sin \theta$ $5=2 \times 5 \sin \theta$ $5=10 \sin \theta$ $\sin \theta=\frac{1}{2}=\sin \frac{\pi}{6}$ $\theta=\pi / 6$
SRMJEEE - 2012
Moving Charges & Magnetism
153825
A current carrying straight wire is kept along the axis of a circular loop carrying a current. The straight wire
1 Will exert an inward force on the circular loop
2 Will exert an outward force on the circular loop
3 Will exert a force on the circular loop parallel to itself
4 Will not exert any force on the circular loop
Explanation:
D The magnetic force on a wire carrying an electric current $\mathrm{i}$ is given as $\overrightarrow{\mathrm{F}}=\mathrm{i}(\vec{l} \times \overrightarrow{\mathrm{B}})$ Where, $l=$ Length of the wire $\mathrm{B}=$ Magnetic field acting on wire If a current carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right hand thumb rule, the magnetic field due to wire on the current carrying loop will be along its circumference, which contains a current element $i \overrightarrow{\mathrm{d} l}$. So, the cross product will be $(\vec{l} \times \overrightarrow{\mathrm{B}})=0$ $\therefore \quad \overrightarrow{\mathrm{F}}=0$ So, the straight wire will not exert any force on the circular loop.
153820
For the magnetic field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be
1 $0^{\circ}$
2 $90^{\circ}$
3 $180^{\circ}$
4 $45^{\circ}$
Explanation:
B The magnetic field due to small element of conductor of length $\mathrm{d} l$ is given by $\mathrm{dB}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{Id} l \sin \theta}{\mathrm{r}^{2}}$ This value will be maximum when, $\sin \theta=1$ $\sin \theta=\sin 90^{\circ}$ $\therefore \quad \theta=90^{\circ}$ Moving Charges \& Magnetism
J and K CET- 2005
Moving Charges & Magnetism
153821
A bar magnet has a magnetic moment of 200 $\mathrm{Am}^{2}$. The magnet is suspended in a magnetic field of $0.30 \mathrm{NA}^{-1} \mathrm{~m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$, will be
153824
A magnet of moment $2 \mathrm{~A}-\mathrm{m}^{2}$ is placed in a uniform magnetic field of $5 \mathrm{~Wb} / \mathrm{m}^{2}$. If the magnet experiences a torque of $5 \mathrm{~N}$, then the angle between the direction of magnetic field and magnet is
1 $\frac{\pi}{6}$
2 $\frac{\pi}{4}$
3 $\frac{\pi}{3}$
4 $\frac{\pi}{2}$
Explanation:
A Given that , $\mathrm{M}=2 \mathrm{~A}-\mathrm{m}^{2}$ $\tau=5 \mathrm{~N} \quad \mathrm{~B}=5 \mathrm{~Wb} / \mathrm{m}^{2}$ We know that, Torque $\tau=\mathrm{MB} \sin \theta$ $5=2 \times 5 \sin \theta$ $5=10 \sin \theta$ $\sin \theta=\frac{1}{2}=\sin \frac{\pi}{6}$ $\theta=\pi / 6$
SRMJEEE - 2012
Moving Charges & Magnetism
153825
A current carrying straight wire is kept along the axis of a circular loop carrying a current. The straight wire
1 Will exert an inward force on the circular loop
2 Will exert an outward force on the circular loop
3 Will exert a force on the circular loop parallel to itself
4 Will not exert any force on the circular loop
Explanation:
D The magnetic force on a wire carrying an electric current $\mathrm{i}$ is given as $\overrightarrow{\mathrm{F}}=\mathrm{i}(\vec{l} \times \overrightarrow{\mathrm{B}})$ Where, $l=$ Length of the wire $\mathrm{B}=$ Magnetic field acting on wire If a current carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right hand thumb rule, the magnetic field due to wire on the current carrying loop will be along its circumference, which contains a current element $i \overrightarrow{\mathrm{d} l}$. So, the cross product will be $(\vec{l} \times \overrightarrow{\mathrm{B}})=0$ $\therefore \quad \overrightarrow{\mathrm{F}}=0$ So, the straight wire will not exert any force on the circular loop.
