153305
An electric field of $1500 \mathrm{~V} / \mathrm{m}$ and a magnetic field of 0.40 weber/metre ${ }^{2}$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is :
1 $1.6 \times 10^{15} \mathrm{~m} / \mathrm{s}$
2 $6 \times 10^{-16} \mathrm{~m} / \mathrm{s}$
3 $3.75 \times 10^{3} \mathrm{~m} / \mathrm{s}$
4 $3.75 \times 10^{2} \mathrm{~m} / \mathrm{s}$
Explanation:
C Given that, $\mathrm{E}=1500 \mathrm{~V} / \mathrm{m}, \mathrm{B}=0.40 \mathrm{~Wb} / \mathrm{m}^{2}$ We know that, $\mathrm{F}_{\mathrm{m}}=\mathrm{F}_{\mathrm{e}}$ $\text { evB }=\text { e.E }$ Minimum speed of electron along the straight line, $v =\frac{E}{B}$ $v =\frac{1500}{0.40}$ $v =3750$ $v =3.75 \times 10^{3} \mathrm{~m} / \mathrm{s}$
Karnataka CET-2005
Moving Charges & Magnetism
153306
Three long straight wires $A, B$ and $C$ are carrying currents as shown in figure. Then the resultant force on $B$ is directed:
1 perpendicular to the plane of paper and outward
2 perpendicular to the plane of paper and inward
3 towards A
4 towards C
Explanation:
D The wire B has a force of attraction from both wires $\mathrm{A}$ and $\mathrm{C}$. The current in all the wire is in same direction. The force between $\mathrm{A}$ and $\mathrm{B}$, $\mathrm{F}_{\mathrm{AB}}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \times 1 \times 2}{\mathrm{~d}}$ $\mathrm{~F}_{\mathrm{AB}}=\frac{4 \mu_{0}}{4 \pi \mathrm{d}}=\frac{\mu_{0}}{\pi \mathrm{d}}$ $\mathrm{F}_{\mathrm{BC}}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \times 2 \times 3}{\mathrm{~d}}$ $\mathrm{F}_{\mathrm{BC}}=\frac{12 \mu_{0}}{4 \pi \mathrm{d}}=\frac{3 \mu_{0}}{\pi \mathrm{d}}$ So, it is clear that, $\mathrm{F}_{\mathrm{BC}}>\mathrm{F}_{\mathrm{AB}}$ Then the resultant force on B is directed towards C.
Karnataka CET-2004
Moving Charges & Magnetism
153308
The magnetic fields at two points on the axis of a circular coil at a distance of $0.05 \mathrm{~m}$ and $0.2 \mathrm{~m}$ from the centre are in the ratio $8: 1$. The radius of the coil is :
1 $0.15 \mathrm{~m}$
2 $0.2 \mathrm{~m}$
3 $1.0 \mathrm{~m}$
4 $0.1 \mathrm{~m}$
Explanation:
D Given that, Distance of points on axis $\mathrm{x}_{1}=0.05, \quad \mathrm{x}_{2}=0.2 \mathrm{~m}$ Ratio of magnetic field. $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{8}{1}$ The magnetic field at the first point $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{Ir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}$ And the magnetic field at the second point $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{Ir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\frac{2\left(\mathrm{r}_{0} \mathrm{Ir}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}{\mu_{0} \mathrm{Ir}^{2}}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}{\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}$ Now squaring both sides, we have $\left(\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}\right)^{2}=\frac{\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3}}{\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3}}$ $\left(\frac{8}{1}\right)^{2 / 3}=\frac{\left(\mathrm{r}^{2}+(0.2)^{2}\right)}{\left(\mathrm{r}^{2}+(0.05)^{2}\right)}$ $4=\frac{r^{2}+0.04}{r^{2}+0.0025}$ $4\left(r^{2}+0.0025\right)=r^{2}+0.04$ $3 r^{2}=0.03$ $r^{2}=0.01$ $r=0.1 \mathrm{~m}$
Karnataka CET-2001
Moving Charges & Magnetism
153312
Graph of force per unit length between two long parallel current carrying conductors and the distance between them is
1 straight line
2 parabola
3 ellipse
4 rectangular hyperbola
Explanation:
D Force per unit length between two long parallel current carrying conductors. $\frac{\mathrm{F}}{l}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{i}_{1} \mathrm{i}_{2}}{\mathrm{y}}$ (Where $\mathrm{y}=$ distance between the two conductor) $\frac{\mathrm{F}}{l} \propto \frac{1}{\mathrm{y}}$ It is the form of rectangular hyperbola.
153305
An electric field of $1500 \mathrm{~V} / \mathrm{m}$ and a magnetic field of 0.40 weber/metre ${ }^{2}$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is :
1 $1.6 \times 10^{15} \mathrm{~m} / \mathrm{s}$
2 $6 \times 10^{-16} \mathrm{~m} / \mathrm{s}$
3 $3.75 \times 10^{3} \mathrm{~m} / \mathrm{s}$
4 $3.75 \times 10^{2} \mathrm{~m} / \mathrm{s}$
Explanation:
C Given that, $\mathrm{E}=1500 \mathrm{~V} / \mathrm{m}, \mathrm{B}=0.40 \mathrm{~Wb} / \mathrm{m}^{2}$ We know that, $\mathrm{F}_{\mathrm{m}}=\mathrm{F}_{\mathrm{e}}$ $\text { evB }=\text { e.E }$ Minimum speed of electron along the straight line, $v =\frac{E}{B}$ $v =\frac{1500}{0.40}$ $v =3750$ $v =3.75 \times 10^{3} \mathrm{~m} / \mathrm{s}$
Karnataka CET-2005
Moving Charges & Magnetism
153306
Three long straight wires $A, B$ and $C$ are carrying currents as shown in figure. Then the resultant force on $B$ is directed:
1 perpendicular to the plane of paper and outward
2 perpendicular to the plane of paper and inward
3 towards A
4 towards C
Explanation:
D The wire B has a force of attraction from both wires $\mathrm{A}$ and $\mathrm{C}$. The current in all the wire is in same direction. The force between $\mathrm{A}$ and $\mathrm{B}$, $\mathrm{F}_{\mathrm{AB}}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \times 1 \times 2}{\mathrm{~d}}$ $\mathrm{~F}_{\mathrm{AB}}=\frac{4 \mu_{0}}{4 \pi \mathrm{d}}=\frac{\mu_{0}}{\pi \mathrm{d}}$ $\mathrm{F}_{\mathrm{BC}}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \times 2 \times 3}{\mathrm{~d}}$ $\mathrm{F}_{\mathrm{BC}}=\frac{12 \mu_{0}}{4 \pi \mathrm{d}}=\frac{3 \mu_{0}}{\pi \mathrm{d}}$ So, it is clear that, $\mathrm{F}_{\mathrm{BC}}>\mathrm{F}_{\mathrm{AB}}$ Then the resultant force on B is directed towards C.
Karnataka CET-2004
Moving Charges & Magnetism
153308
The magnetic fields at two points on the axis of a circular coil at a distance of $0.05 \mathrm{~m}$ and $0.2 \mathrm{~m}$ from the centre are in the ratio $8: 1$. The radius of the coil is :
1 $0.15 \mathrm{~m}$
2 $0.2 \mathrm{~m}$
3 $1.0 \mathrm{~m}$
4 $0.1 \mathrm{~m}$
Explanation:
D Given that, Distance of points on axis $\mathrm{x}_{1}=0.05, \quad \mathrm{x}_{2}=0.2 \mathrm{~m}$ Ratio of magnetic field. $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{8}{1}$ The magnetic field at the first point $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{Ir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}$ And the magnetic field at the second point $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{Ir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\frac{2\left(\mathrm{r}_{0} \mathrm{Ir}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}{\mu_{0} \mathrm{Ir}^{2}}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}{\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}$ Now squaring both sides, we have $\left(\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}\right)^{2}=\frac{\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3}}{\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3}}$ $\left(\frac{8}{1}\right)^{2 / 3}=\frac{\left(\mathrm{r}^{2}+(0.2)^{2}\right)}{\left(\mathrm{r}^{2}+(0.05)^{2}\right)}$ $4=\frac{r^{2}+0.04}{r^{2}+0.0025}$ $4\left(r^{2}+0.0025\right)=r^{2}+0.04$ $3 r^{2}=0.03$ $r^{2}=0.01$ $r=0.1 \mathrm{~m}$
Karnataka CET-2001
Moving Charges & Magnetism
153312
Graph of force per unit length between two long parallel current carrying conductors and the distance between them is
1 straight line
2 parabola
3 ellipse
4 rectangular hyperbola
Explanation:
D Force per unit length between two long parallel current carrying conductors. $\frac{\mathrm{F}}{l}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{i}_{1} \mathrm{i}_{2}}{\mathrm{y}}$ (Where $\mathrm{y}=$ distance between the two conductor) $\frac{\mathrm{F}}{l} \propto \frac{1}{\mathrm{y}}$ It is the form of rectangular hyperbola.
153305
An electric field of $1500 \mathrm{~V} / \mathrm{m}$ and a magnetic field of 0.40 weber/metre ${ }^{2}$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is :
1 $1.6 \times 10^{15} \mathrm{~m} / \mathrm{s}$
2 $6 \times 10^{-16} \mathrm{~m} / \mathrm{s}$
3 $3.75 \times 10^{3} \mathrm{~m} / \mathrm{s}$
4 $3.75 \times 10^{2} \mathrm{~m} / \mathrm{s}$
Explanation:
C Given that, $\mathrm{E}=1500 \mathrm{~V} / \mathrm{m}, \mathrm{B}=0.40 \mathrm{~Wb} / \mathrm{m}^{2}$ We know that, $\mathrm{F}_{\mathrm{m}}=\mathrm{F}_{\mathrm{e}}$ $\text { evB }=\text { e.E }$ Minimum speed of electron along the straight line, $v =\frac{E}{B}$ $v =\frac{1500}{0.40}$ $v =3750$ $v =3.75 \times 10^{3} \mathrm{~m} / \mathrm{s}$
Karnataka CET-2005
Moving Charges & Magnetism
153306
Three long straight wires $A, B$ and $C$ are carrying currents as shown in figure. Then the resultant force on $B$ is directed:
1 perpendicular to the plane of paper and outward
2 perpendicular to the plane of paper and inward
3 towards A
4 towards C
Explanation:
D The wire B has a force of attraction from both wires $\mathrm{A}$ and $\mathrm{C}$. The current in all the wire is in same direction. The force between $\mathrm{A}$ and $\mathrm{B}$, $\mathrm{F}_{\mathrm{AB}}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \times 1 \times 2}{\mathrm{~d}}$ $\mathrm{~F}_{\mathrm{AB}}=\frac{4 \mu_{0}}{4 \pi \mathrm{d}}=\frac{\mu_{0}}{\pi \mathrm{d}}$ $\mathrm{F}_{\mathrm{BC}}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \times 2 \times 3}{\mathrm{~d}}$ $\mathrm{F}_{\mathrm{BC}}=\frac{12 \mu_{0}}{4 \pi \mathrm{d}}=\frac{3 \mu_{0}}{\pi \mathrm{d}}$ So, it is clear that, $\mathrm{F}_{\mathrm{BC}}>\mathrm{F}_{\mathrm{AB}}$ Then the resultant force on B is directed towards C.
Karnataka CET-2004
Moving Charges & Magnetism
153308
The magnetic fields at two points on the axis of a circular coil at a distance of $0.05 \mathrm{~m}$ and $0.2 \mathrm{~m}$ from the centre are in the ratio $8: 1$. The radius of the coil is :
1 $0.15 \mathrm{~m}$
2 $0.2 \mathrm{~m}$
3 $1.0 \mathrm{~m}$
4 $0.1 \mathrm{~m}$
Explanation:
D Given that, Distance of points on axis $\mathrm{x}_{1}=0.05, \quad \mathrm{x}_{2}=0.2 \mathrm{~m}$ Ratio of magnetic field. $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{8}{1}$ The magnetic field at the first point $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{Ir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}$ And the magnetic field at the second point $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{Ir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\frac{2\left(\mathrm{r}_{0} \mathrm{Ir}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}{\mu_{0} \mathrm{Ir}^{2}}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}{\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}$ Now squaring both sides, we have $\left(\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}\right)^{2}=\frac{\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3}}{\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3}}$ $\left(\frac{8}{1}\right)^{2 / 3}=\frac{\left(\mathrm{r}^{2}+(0.2)^{2}\right)}{\left(\mathrm{r}^{2}+(0.05)^{2}\right)}$ $4=\frac{r^{2}+0.04}{r^{2}+0.0025}$ $4\left(r^{2}+0.0025\right)=r^{2}+0.04$ $3 r^{2}=0.03$ $r^{2}=0.01$ $r=0.1 \mathrm{~m}$
Karnataka CET-2001
Moving Charges & Magnetism
153312
Graph of force per unit length between two long parallel current carrying conductors and the distance between them is
1 straight line
2 parabola
3 ellipse
4 rectangular hyperbola
Explanation:
D Force per unit length between two long parallel current carrying conductors. $\frac{\mathrm{F}}{l}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{i}_{1} \mathrm{i}_{2}}{\mathrm{y}}$ (Where $\mathrm{y}=$ distance between the two conductor) $\frac{\mathrm{F}}{l} \propto \frac{1}{\mathrm{y}}$ It is the form of rectangular hyperbola.
153305
An electric field of $1500 \mathrm{~V} / \mathrm{m}$ and a magnetic field of 0.40 weber/metre ${ }^{2}$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is :
1 $1.6 \times 10^{15} \mathrm{~m} / \mathrm{s}$
2 $6 \times 10^{-16} \mathrm{~m} / \mathrm{s}$
3 $3.75 \times 10^{3} \mathrm{~m} / \mathrm{s}$
4 $3.75 \times 10^{2} \mathrm{~m} / \mathrm{s}$
Explanation:
C Given that, $\mathrm{E}=1500 \mathrm{~V} / \mathrm{m}, \mathrm{B}=0.40 \mathrm{~Wb} / \mathrm{m}^{2}$ We know that, $\mathrm{F}_{\mathrm{m}}=\mathrm{F}_{\mathrm{e}}$ $\text { evB }=\text { e.E }$ Minimum speed of electron along the straight line, $v =\frac{E}{B}$ $v =\frac{1500}{0.40}$ $v =3750$ $v =3.75 \times 10^{3} \mathrm{~m} / \mathrm{s}$
Karnataka CET-2005
Moving Charges & Magnetism
153306
Three long straight wires $A, B$ and $C$ are carrying currents as shown in figure. Then the resultant force on $B$ is directed:
1 perpendicular to the plane of paper and outward
2 perpendicular to the plane of paper and inward
3 towards A
4 towards C
Explanation:
D The wire B has a force of attraction from both wires $\mathrm{A}$ and $\mathrm{C}$. The current in all the wire is in same direction. The force between $\mathrm{A}$ and $\mathrm{B}$, $\mathrm{F}_{\mathrm{AB}}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \times 1 \times 2}{\mathrm{~d}}$ $\mathrm{~F}_{\mathrm{AB}}=\frac{4 \mu_{0}}{4 \pi \mathrm{d}}=\frac{\mu_{0}}{\pi \mathrm{d}}$ $\mathrm{F}_{\mathrm{BC}}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \times 2 \times 3}{\mathrm{~d}}$ $\mathrm{F}_{\mathrm{BC}}=\frac{12 \mu_{0}}{4 \pi \mathrm{d}}=\frac{3 \mu_{0}}{\pi \mathrm{d}}$ So, it is clear that, $\mathrm{F}_{\mathrm{BC}}>\mathrm{F}_{\mathrm{AB}}$ Then the resultant force on B is directed towards C.
Karnataka CET-2004
Moving Charges & Magnetism
153308
The magnetic fields at two points on the axis of a circular coil at a distance of $0.05 \mathrm{~m}$ and $0.2 \mathrm{~m}$ from the centre are in the ratio $8: 1$. The radius of the coil is :
1 $0.15 \mathrm{~m}$
2 $0.2 \mathrm{~m}$
3 $1.0 \mathrm{~m}$
4 $0.1 \mathrm{~m}$
Explanation:
D Given that, Distance of points on axis $\mathrm{x}_{1}=0.05, \quad \mathrm{x}_{2}=0.2 \mathrm{~m}$ Ratio of magnetic field. $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{8}{1}$ The magnetic field at the first point $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{Ir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}$ And the magnetic field at the second point $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{Ir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\frac{2\left(\mathrm{r}_{0} \mathrm{Ir}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}{\mu_{0} \mathrm{Ir}^{2}}}{2\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3 / 2}}{\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3 / 2}}$ Now squaring both sides, we have $\left(\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}\right)^{2}=\frac{\left(\mathrm{r}^{2}+\mathrm{x}_{2}^{2}\right)^{3}}{\left(\mathrm{r}^{2}+\mathrm{x}_{1}^{2}\right)^{3}}$ $\left(\frac{8}{1}\right)^{2 / 3}=\frac{\left(\mathrm{r}^{2}+(0.2)^{2}\right)}{\left(\mathrm{r}^{2}+(0.05)^{2}\right)}$ $4=\frac{r^{2}+0.04}{r^{2}+0.0025}$ $4\left(r^{2}+0.0025\right)=r^{2}+0.04$ $3 r^{2}=0.03$ $r^{2}=0.01$ $r=0.1 \mathrm{~m}$
Karnataka CET-2001
Moving Charges & Magnetism
153312
Graph of force per unit length between two long parallel current carrying conductors and the distance between them is
1 straight line
2 parabola
3 ellipse
4 rectangular hyperbola
Explanation:
D Force per unit length between two long parallel current carrying conductors. $\frac{\mathrm{F}}{l}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{i}_{1} \mathrm{i}_{2}}{\mathrm{y}}$ (Where $\mathrm{y}=$ distance between the two conductor) $\frac{\mathrm{F}}{l} \propto \frac{1}{\mathrm{y}}$ It is the form of rectangular hyperbola.
