NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153355
A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by
1 $\frac{\mathrm{B}^{2}}{2 \mathrm{VE}^{2}}$
2 $\frac{2 \mathrm{VB}^{2}}{\mathrm{E}^{2}}$
3 $\frac{2 \mathrm{VE}^{2}}{\mathrm{~B}^{2}}$
4 $\frac{E^{2}}{2 \mathrm{VB}^{2}}$
Explanation:
D When the electron beam is not deflected then $\mathrm{F}_{\mathrm{m}}=\mathrm{F}_{\mathrm{e}}$ $\mathrm{Bev}=\mathrm{Ee}$ $v=\frac{E}{B}$ The law of conservation of energy, $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$ $\frac{1}{2} \mathrm{~m} \times\left(\frac{\mathrm{E}}{\mathrm{B}}\right)^{2}=\mathrm{eV}$ $\frac{\mathrm{e}}{\mathrm{m}}=\frac{\mathrm{E}^{2}}{2 \mathrm{VB}^{2}}$
AIPMT-2010
Moving Charges & Magnetism
153356
A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic induction at their centres will be
1 $2: 1$
2 $1: 4$
3 $4: 1$
4 $1: 2$
Explanation:
B Let the radii be $\mathrm{r}_{1}$ and $\mathrm{r}_{2}$ respectively. Since there are two turns of radius $r_{2}$ $r_{1}=2 r_{2}$ Magnetic field $\mathrm{B}$ at the centre of the coil radius $\mathrm{r}_{1}$ $B_{1}=\frac{\mu_{0} i}{2 r_{1}}=\frac{\mu_{0} i}{4 r_{2}}$ Magnetic field $B$ at the centre of the coil radius $r_{2}$ $\mathrm{B}_{2}=\frac{2 \times \mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{r}_{2}}$ On dividing equation (ii) by (i), we get $\frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}}=\frac{\frac{2 \times \mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{r}_{2}}}{\frac{\mu_{\mathrm{o}} \mathrm{i}}{4 \mathrm{r}_{2}}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{4} \text { or } 1: 4$
AIPMT-1998
Moving Charges & Magnetism
153357
At what distance from a long straight wire carrying a current of $12 \mathrm{~A}$ will the magnetic field be equal to $3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}$ ?
1 $8 \times 10^{-2} \mathrm{~m}$
2 $12 \times 10^{-2} \mathrm{~m}$
3 $18 \times 10^{-2} \mathrm{~m}$
4 $24 \times 10^{-2} \mathrm{~m}$
Explanation:
A Given, $\mathrm{B}=3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}$ $\mathrm{i}=12 \mathrm{~A}$ The magnetic field at distance $r$ from the long current carrying straight wire is $B =\frac{\mu_{0} i}{2 \pi r}$ $r =\frac{\mu_{0} i}{2 \pi B}$ $=\frac{4 \pi \times 10^{-7} \times 12}{2 \pi \times 3 \times 10^{-5}}$ $r =8 \times 10^{-2} \mathrm{~m}$
AIPMT-1995
Moving Charges & Magnetism
153358
The magnetic field at a distance $r$ from a long wire carrying current $I$ is $0.4 \mathrm{~T}$. The magnetic field at a distance $2 r$ is
1 $0.2 \mathrm{~T}$
2 $0.8 \mathrm{~T}$
3 $0.1 \mathrm{~T}$
4 $1.6 \mathrm{~T}$
Explanation:
A Given, $\mathrm{B}_{1}=0.4 \mathrm{~T}$ Magnetic field at a distance $r$ for a long wire carrying current i, $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{2 \mathrm{r}}{\mathrm{r}}$ So, $\quad \mathrm{B}_{2}=\frac{\mathrm{B}_{1}}{2}=\frac{0.4}{2}$ $\mathrm{B}_{2}=0.2 \mathrm{~T}$
153355
A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by
1 $\frac{\mathrm{B}^{2}}{2 \mathrm{VE}^{2}}$
2 $\frac{2 \mathrm{VB}^{2}}{\mathrm{E}^{2}}$
3 $\frac{2 \mathrm{VE}^{2}}{\mathrm{~B}^{2}}$
4 $\frac{E^{2}}{2 \mathrm{VB}^{2}}$
Explanation:
D When the electron beam is not deflected then $\mathrm{F}_{\mathrm{m}}=\mathrm{F}_{\mathrm{e}}$ $\mathrm{Bev}=\mathrm{Ee}$ $v=\frac{E}{B}$ The law of conservation of energy, $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$ $\frac{1}{2} \mathrm{~m} \times\left(\frac{\mathrm{E}}{\mathrm{B}}\right)^{2}=\mathrm{eV}$ $\frac{\mathrm{e}}{\mathrm{m}}=\frac{\mathrm{E}^{2}}{2 \mathrm{VB}^{2}}$
AIPMT-2010
Moving Charges & Magnetism
153356
A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic induction at their centres will be
1 $2: 1$
2 $1: 4$
3 $4: 1$
4 $1: 2$
Explanation:
B Let the radii be $\mathrm{r}_{1}$ and $\mathrm{r}_{2}$ respectively. Since there are two turns of radius $r_{2}$ $r_{1}=2 r_{2}$ Magnetic field $\mathrm{B}$ at the centre of the coil radius $\mathrm{r}_{1}$ $B_{1}=\frac{\mu_{0} i}{2 r_{1}}=\frac{\mu_{0} i}{4 r_{2}}$ Magnetic field $B$ at the centre of the coil radius $r_{2}$ $\mathrm{B}_{2}=\frac{2 \times \mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{r}_{2}}$ On dividing equation (ii) by (i), we get $\frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}}=\frac{\frac{2 \times \mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{r}_{2}}}{\frac{\mu_{\mathrm{o}} \mathrm{i}}{4 \mathrm{r}_{2}}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{4} \text { or } 1: 4$
AIPMT-1998
Moving Charges & Magnetism
153357
At what distance from a long straight wire carrying a current of $12 \mathrm{~A}$ will the magnetic field be equal to $3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}$ ?
1 $8 \times 10^{-2} \mathrm{~m}$
2 $12 \times 10^{-2} \mathrm{~m}$
3 $18 \times 10^{-2} \mathrm{~m}$
4 $24 \times 10^{-2} \mathrm{~m}$
Explanation:
A Given, $\mathrm{B}=3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}$ $\mathrm{i}=12 \mathrm{~A}$ The magnetic field at distance $r$ from the long current carrying straight wire is $B =\frac{\mu_{0} i}{2 \pi r}$ $r =\frac{\mu_{0} i}{2 \pi B}$ $=\frac{4 \pi \times 10^{-7} \times 12}{2 \pi \times 3 \times 10^{-5}}$ $r =8 \times 10^{-2} \mathrm{~m}$
AIPMT-1995
Moving Charges & Magnetism
153358
The magnetic field at a distance $r$ from a long wire carrying current $I$ is $0.4 \mathrm{~T}$. The magnetic field at a distance $2 r$ is
1 $0.2 \mathrm{~T}$
2 $0.8 \mathrm{~T}$
3 $0.1 \mathrm{~T}$
4 $1.6 \mathrm{~T}$
Explanation:
A Given, $\mathrm{B}_{1}=0.4 \mathrm{~T}$ Magnetic field at a distance $r$ for a long wire carrying current i, $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{2 \mathrm{r}}{\mathrm{r}}$ So, $\quad \mathrm{B}_{2}=\frac{\mathrm{B}_{1}}{2}=\frac{0.4}{2}$ $\mathrm{B}_{2}=0.2 \mathrm{~T}$
153355
A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by
1 $\frac{\mathrm{B}^{2}}{2 \mathrm{VE}^{2}}$
2 $\frac{2 \mathrm{VB}^{2}}{\mathrm{E}^{2}}$
3 $\frac{2 \mathrm{VE}^{2}}{\mathrm{~B}^{2}}$
4 $\frac{E^{2}}{2 \mathrm{VB}^{2}}$
Explanation:
D When the electron beam is not deflected then $\mathrm{F}_{\mathrm{m}}=\mathrm{F}_{\mathrm{e}}$ $\mathrm{Bev}=\mathrm{Ee}$ $v=\frac{E}{B}$ The law of conservation of energy, $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$ $\frac{1}{2} \mathrm{~m} \times\left(\frac{\mathrm{E}}{\mathrm{B}}\right)^{2}=\mathrm{eV}$ $\frac{\mathrm{e}}{\mathrm{m}}=\frac{\mathrm{E}^{2}}{2 \mathrm{VB}^{2}}$
AIPMT-2010
Moving Charges & Magnetism
153356
A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic induction at their centres will be
1 $2: 1$
2 $1: 4$
3 $4: 1$
4 $1: 2$
Explanation:
B Let the radii be $\mathrm{r}_{1}$ and $\mathrm{r}_{2}$ respectively. Since there are two turns of radius $r_{2}$ $r_{1}=2 r_{2}$ Magnetic field $\mathrm{B}$ at the centre of the coil radius $\mathrm{r}_{1}$ $B_{1}=\frac{\mu_{0} i}{2 r_{1}}=\frac{\mu_{0} i}{4 r_{2}}$ Magnetic field $B$ at the centre of the coil radius $r_{2}$ $\mathrm{B}_{2}=\frac{2 \times \mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{r}_{2}}$ On dividing equation (ii) by (i), we get $\frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}}=\frac{\frac{2 \times \mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{r}_{2}}}{\frac{\mu_{\mathrm{o}} \mathrm{i}}{4 \mathrm{r}_{2}}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{4} \text { or } 1: 4$
AIPMT-1998
Moving Charges & Magnetism
153357
At what distance from a long straight wire carrying a current of $12 \mathrm{~A}$ will the magnetic field be equal to $3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}$ ?
1 $8 \times 10^{-2} \mathrm{~m}$
2 $12 \times 10^{-2} \mathrm{~m}$
3 $18 \times 10^{-2} \mathrm{~m}$
4 $24 \times 10^{-2} \mathrm{~m}$
Explanation:
A Given, $\mathrm{B}=3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}$ $\mathrm{i}=12 \mathrm{~A}$ The magnetic field at distance $r$ from the long current carrying straight wire is $B =\frac{\mu_{0} i}{2 \pi r}$ $r =\frac{\mu_{0} i}{2 \pi B}$ $=\frac{4 \pi \times 10^{-7} \times 12}{2 \pi \times 3 \times 10^{-5}}$ $r =8 \times 10^{-2} \mathrm{~m}$
AIPMT-1995
Moving Charges & Magnetism
153358
The magnetic field at a distance $r$ from a long wire carrying current $I$ is $0.4 \mathrm{~T}$. The magnetic field at a distance $2 r$ is
1 $0.2 \mathrm{~T}$
2 $0.8 \mathrm{~T}$
3 $0.1 \mathrm{~T}$
4 $1.6 \mathrm{~T}$
Explanation:
A Given, $\mathrm{B}_{1}=0.4 \mathrm{~T}$ Magnetic field at a distance $r$ for a long wire carrying current i, $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{2 \mathrm{r}}{\mathrm{r}}$ So, $\quad \mathrm{B}_{2}=\frac{\mathrm{B}_{1}}{2}=\frac{0.4}{2}$ $\mathrm{B}_{2}=0.2 \mathrm{~T}$
153355
A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by
1 $\frac{\mathrm{B}^{2}}{2 \mathrm{VE}^{2}}$
2 $\frac{2 \mathrm{VB}^{2}}{\mathrm{E}^{2}}$
3 $\frac{2 \mathrm{VE}^{2}}{\mathrm{~B}^{2}}$
4 $\frac{E^{2}}{2 \mathrm{VB}^{2}}$
Explanation:
D When the electron beam is not deflected then $\mathrm{F}_{\mathrm{m}}=\mathrm{F}_{\mathrm{e}}$ $\mathrm{Bev}=\mathrm{Ee}$ $v=\frac{E}{B}$ The law of conservation of energy, $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$ $\frac{1}{2} \mathrm{~m} \times\left(\frac{\mathrm{E}}{\mathrm{B}}\right)^{2}=\mathrm{eV}$ $\frac{\mathrm{e}}{\mathrm{m}}=\frac{\mathrm{E}^{2}}{2 \mathrm{VB}^{2}}$
AIPMT-2010
Moving Charges & Magnetism
153356
A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic induction at their centres will be
1 $2: 1$
2 $1: 4$
3 $4: 1$
4 $1: 2$
Explanation:
B Let the radii be $\mathrm{r}_{1}$ and $\mathrm{r}_{2}$ respectively. Since there are two turns of radius $r_{2}$ $r_{1}=2 r_{2}$ Magnetic field $\mathrm{B}$ at the centre of the coil radius $\mathrm{r}_{1}$ $B_{1}=\frac{\mu_{0} i}{2 r_{1}}=\frac{\mu_{0} i}{4 r_{2}}$ Magnetic field $B$ at the centre of the coil radius $r_{2}$ $\mathrm{B}_{2}=\frac{2 \times \mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{r}_{2}}$ On dividing equation (ii) by (i), we get $\frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}}=\frac{\frac{2 \times \mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{r}_{2}}}{\frac{\mu_{\mathrm{o}} \mathrm{i}}{4 \mathrm{r}_{2}}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{4} \text { or } 1: 4$
AIPMT-1998
Moving Charges & Magnetism
153357
At what distance from a long straight wire carrying a current of $12 \mathrm{~A}$ will the magnetic field be equal to $3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}$ ?
1 $8 \times 10^{-2} \mathrm{~m}$
2 $12 \times 10^{-2} \mathrm{~m}$
3 $18 \times 10^{-2} \mathrm{~m}$
4 $24 \times 10^{-2} \mathrm{~m}$
Explanation:
A Given, $\mathrm{B}=3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}$ $\mathrm{i}=12 \mathrm{~A}$ The magnetic field at distance $r$ from the long current carrying straight wire is $B =\frac{\mu_{0} i}{2 \pi r}$ $r =\frac{\mu_{0} i}{2 \pi B}$ $=\frac{4 \pi \times 10^{-7} \times 12}{2 \pi \times 3 \times 10^{-5}}$ $r =8 \times 10^{-2} \mathrm{~m}$
AIPMT-1995
Moving Charges & Magnetism
153358
The magnetic field at a distance $r$ from a long wire carrying current $I$ is $0.4 \mathrm{~T}$. The magnetic field at a distance $2 r$ is
1 $0.2 \mathrm{~T}$
2 $0.8 \mathrm{~T}$
3 $0.1 \mathrm{~T}$
4 $1.6 \mathrm{~T}$
Explanation:
A Given, $\mathrm{B}_{1}=0.4 \mathrm{~T}$ Magnetic field at a distance $r$ for a long wire carrying current i, $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{2 \mathrm{r}}{\mathrm{r}}$ So, $\quad \mathrm{B}_{2}=\frac{\mathrm{B}_{1}}{2}=\frac{0.4}{2}$ $\mathrm{B}_{2}=0.2 \mathrm{~T}$
