NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153283
In a circular coil (1) of radius $R$, current $I$ is flowing and in another coil (2) or radius $2 R$ a current 2I is flowing, then the ratio of the magnetic fields produced by the two coils is-
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $3: 1$
Explanation:
A As we know that, Magnetic field in a circular coil, $\mathrm{B}=\frac{\mu_{0}}{2}\left(\frac{\mathrm{I}}{\mathrm{R}}\right)$ $\therefore \quad \mathrm{B} \propto \frac{\mathrm{I}}{\mathrm{R}}$ Given that, $\mathrm{I}_{1}=\mathrm{I}$ and $\mathrm{I}_{2}=2 \mathrm{I}$ $\mathrm{R}_{1}=\mathrm{R} \text { and } \mathrm{R}_{2}=2 \mathrm{R}$ Ratio of magnetic field, $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}} \times \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mathrm{I}}{2 \mathrm{I}} \times \frac{2 \mathrm{R}}{\mathrm{R}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{1}$ $\mathrm{~B}_{1}: \mathrm{B}_{2}=1: 1$
BCECE-2014
Moving Charges & Magnetism
153284
A horizontal overhead power line carries acurrent of $90 \mathrm{~A}$ in east to west direction. Magnitude of magnetic field due to the current $1.5 \mathrm{~m}$ below the line is-
1 $1.2 \mathrm{~T}$
2 $1.2 \times 10^{-10} \mathrm{~T}$
3 $0 \mathrm{~T}$
4 $1.2 \times 10^{-5} \mathrm{~T}$
Explanation:
D Given that, $\mathrm{I}=90 \mathrm{~A}, \mathrm{~d}=1.5 \mathrm{~m}$ Magnitude of magnetic field at a distance $d$ due to straight current carrying conductor is given by- $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}$ $\mathrm{B}=\frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$ $\mathrm{~B}=2 \times 10^{-7} \times 60$ $\mathrm{~B}=120 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}=1.2 \times 10^{-5} \mathrm{~T}$
BCECE-2014
Moving Charges & Magnetism
153285
Which of the following graphs shows the variation of magnetic field $B$, with distance from a long current carrying conductor?
1 a
2 b
3 c
4 d
Explanation:
C Magnetic field due to long current carrying conductor at a distance $r$ is- $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ So, the graph between the magnetic field and the distance $r$ is-
COMEDK 2015
Moving Charges & Magnetism
153286
The distance between the poles of a horse shoe magnet is $0.1 \mathrm{~m}$ and its pole strength is $0.01 \mathrm{~A}$ $m$. The induction of magnetic field at a point mid way between the poles will be-
1 Zero
2 $2 \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-6} \mathrm{~T}$
4 $8 \times 10^{-7} \mathrm{~T}$
Explanation:
D Given, $\mathrm{d}=0.1 \mathrm{~m} \quad \mathrm{r}=\frac{0.1}{2}=0.05 \mathrm{~m}$ Pole strength $=\mathrm{m}=0.01 \quad$ A-m Net magnetic field at midpoint $\mathrm{O}$, $\mathrm{B}_{\text {net }}=\mathrm{B}_{\mathrm{N}}+\mathrm{B}_{\mathrm{S}}$ Where, $\mathrm{B}_{\mathrm{N}}=$ Magnitude of magnetic field due to $\mathrm{N}$ Pole $\mathrm{B}_{\mathrm{S}}=$ Magnitude of magnetic field due to $\mathrm{S}$ pole. $\mathrm{B}_{\mathrm{S}}=\mathrm{B}_{\mathrm{N}}=\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}}$ $\mathrm{~B}=\frac{4 \pi \times 10^{-7} \times 0.01}{4 \pi\left(\frac{0.1}{2}\right)^{2}}=4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{\text {net }}=2 \mathrm{~B}=2 \times 4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{\text {net }}=8 \times 10^{-7} \mathrm{~T}$
153283
In a circular coil (1) of radius $R$, current $I$ is flowing and in another coil (2) or radius $2 R$ a current 2I is flowing, then the ratio of the magnetic fields produced by the two coils is-
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $3: 1$
Explanation:
A As we know that, Magnetic field in a circular coil, $\mathrm{B}=\frac{\mu_{0}}{2}\left(\frac{\mathrm{I}}{\mathrm{R}}\right)$ $\therefore \quad \mathrm{B} \propto \frac{\mathrm{I}}{\mathrm{R}}$ Given that, $\mathrm{I}_{1}=\mathrm{I}$ and $\mathrm{I}_{2}=2 \mathrm{I}$ $\mathrm{R}_{1}=\mathrm{R} \text { and } \mathrm{R}_{2}=2 \mathrm{R}$ Ratio of magnetic field, $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}} \times \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mathrm{I}}{2 \mathrm{I}} \times \frac{2 \mathrm{R}}{\mathrm{R}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{1}$ $\mathrm{~B}_{1}: \mathrm{B}_{2}=1: 1$
BCECE-2014
Moving Charges & Magnetism
153284
A horizontal overhead power line carries acurrent of $90 \mathrm{~A}$ in east to west direction. Magnitude of magnetic field due to the current $1.5 \mathrm{~m}$ below the line is-
1 $1.2 \mathrm{~T}$
2 $1.2 \times 10^{-10} \mathrm{~T}$
3 $0 \mathrm{~T}$
4 $1.2 \times 10^{-5} \mathrm{~T}$
Explanation:
D Given that, $\mathrm{I}=90 \mathrm{~A}, \mathrm{~d}=1.5 \mathrm{~m}$ Magnitude of magnetic field at a distance $d$ due to straight current carrying conductor is given by- $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}$ $\mathrm{B}=\frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$ $\mathrm{~B}=2 \times 10^{-7} \times 60$ $\mathrm{~B}=120 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}=1.2 \times 10^{-5} \mathrm{~T}$
BCECE-2014
Moving Charges & Magnetism
153285
Which of the following graphs shows the variation of magnetic field $B$, with distance from a long current carrying conductor?
1 a
2 b
3 c
4 d
Explanation:
C Magnetic field due to long current carrying conductor at a distance $r$ is- $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ So, the graph between the magnetic field and the distance $r$ is-
COMEDK 2015
Moving Charges & Magnetism
153286
The distance between the poles of a horse shoe magnet is $0.1 \mathrm{~m}$ and its pole strength is $0.01 \mathrm{~A}$ $m$. The induction of magnetic field at a point mid way between the poles will be-
1 Zero
2 $2 \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-6} \mathrm{~T}$
4 $8 \times 10^{-7} \mathrm{~T}$
Explanation:
D Given, $\mathrm{d}=0.1 \mathrm{~m} \quad \mathrm{r}=\frac{0.1}{2}=0.05 \mathrm{~m}$ Pole strength $=\mathrm{m}=0.01 \quad$ A-m Net magnetic field at midpoint $\mathrm{O}$, $\mathrm{B}_{\text {net }}=\mathrm{B}_{\mathrm{N}}+\mathrm{B}_{\mathrm{S}}$ Where, $\mathrm{B}_{\mathrm{N}}=$ Magnitude of magnetic field due to $\mathrm{N}$ Pole $\mathrm{B}_{\mathrm{S}}=$ Magnitude of magnetic field due to $\mathrm{S}$ pole. $\mathrm{B}_{\mathrm{S}}=\mathrm{B}_{\mathrm{N}}=\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}}$ $\mathrm{~B}=\frac{4 \pi \times 10^{-7} \times 0.01}{4 \pi\left(\frac{0.1}{2}\right)^{2}}=4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{\text {net }}=2 \mathrm{~B}=2 \times 4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{\text {net }}=8 \times 10^{-7} \mathrm{~T}$
153283
In a circular coil (1) of radius $R$, current $I$ is flowing and in another coil (2) or radius $2 R$ a current 2I is flowing, then the ratio of the magnetic fields produced by the two coils is-
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $3: 1$
Explanation:
A As we know that, Magnetic field in a circular coil, $\mathrm{B}=\frac{\mu_{0}}{2}\left(\frac{\mathrm{I}}{\mathrm{R}}\right)$ $\therefore \quad \mathrm{B} \propto \frac{\mathrm{I}}{\mathrm{R}}$ Given that, $\mathrm{I}_{1}=\mathrm{I}$ and $\mathrm{I}_{2}=2 \mathrm{I}$ $\mathrm{R}_{1}=\mathrm{R} \text { and } \mathrm{R}_{2}=2 \mathrm{R}$ Ratio of magnetic field, $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}} \times \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mathrm{I}}{2 \mathrm{I}} \times \frac{2 \mathrm{R}}{\mathrm{R}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{1}$ $\mathrm{~B}_{1}: \mathrm{B}_{2}=1: 1$
BCECE-2014
Moving Charges & Magnetism
153284
A horizontal overhead power line carries acurrent of $90 \mathrm{~A}$ in east to west direction. Magnitude of magnetic field due to the current $1.5 \mathrm{~m}$ below the line is-
1 $1.2 \mathrm{~T}$
2 $1.2 \times 10^{-10} \mathrm{~T}$
3 $0 \mathrm{~T}$
4 $1.2 \times 10^{-5} \mathrm{~T}$
Explanation:
D Given that, $\mathrm{I}=90 \mathrm{~A}, \mathrm{~d}=1.5 \mathrm{~m}$ Magnitude of magnetic field at a distance $d$ due to straight current carrying conductor is given by- $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}$ $\mathrm{B}=\frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$ $\mathrm{~B}=2 \times 10^{-7} \times 60$ $\mathrm{~B}=120 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}=1.2 \times 10^{-5} \mathrm{~T}$
BCECE-2014
Moving Charges & Magnetism
153285
Which of the following graphs shows the variation of magnetic field $B$, with distance from a long current carrying conductor?
1 a
2 b
3 c
4 d
Explanation:
C Magnetic field due to long current carrying conductor at a distance $r$ is- $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ So, the graph between the magnetic field and the distance $r$ is-
COMEDK 2015
Moving Charges & Magnetism
153286
The distance between the poles of a horse shoe magnet is $0.1 \mathrm{~m}$ and its pole strength is $0.01 \mathrm{~A}$ $m$. The induction of magnetic field at a point mid way between the poles will be-
1 Zero
2 $2 \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-6} \mathrm{~T}$
4 $8 \times 10^{-7} \mathrm{~T}$
Explanation:
D Given, $\mathrm{d}=0.1 \mathrm{~m} \quad \mathrm{r}=\frac{0.1}{2}=0.05 \mathrm{~m}$ Pole strength $=\mathrm{m}=0.01 \quad$ A-m Net magnetic field at midpoint $\mathrm{O}$, $\mathrm{B}_{\text {net }}=\mathrm{B}_{\mathrm{N}}+\mathrm{B}_{\mathrm{S}}$ Where, $\mathrm{B}_{\mathrm{N}}=$ Magnitude of magnetic field due to $\mathrm{N}$ Pole $\mathrm{B}_{\mathrm{S}}=$ Magnitude of magnetic field due to $\mathrm{S}$ pole. $\mathrm{B}_{\mathrm{S}}=\mathrm{B}_{\mathrm{N}}=\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}}$ $\mathrm{~B}=\frac{4 \pi \times 10^{-7} \times 0.01}{4 \pi\left(\frac{0.1}{2}\right)^{2}}=4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{\text {net }}=2 \mathrm{~B}=2 \times 4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{\text {net }}=8 \times 10^{-7} \mathrm{~T}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Moving Charges & Magnetism
153283
In a circular coil (1) of radius $R$, current $I$ is flowing and in another coil (2) or radius $2 R$ a current 2I is flowing, then the ratio of the magnetic fields produced by the two coils is-
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $3: 1$
Explanation:
A As we know that, Magnetic field in a circular coil, $\mathrm{B}=\frac{\mu_{0}}{2}\left(\frac{\mathrm{I}}{\mathrm{R}}\right)$ $\therefore \quad \mathrm{B} \propto \frac{\mathrm{I}}{\mathrm{R}}$ Given that, $\mathrm{I}_{1}=\mathrm{I}$ and $\mathrm{I}_{2}=2 \mathrm{I}$ $\mathrm{R}_{1}=\mathrm{R} \text { and } \mathrm{R}_{2}=2 \mathrm{R}$ Ratio of magnetic field, $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}} \times \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mathrm{I}}{2 \mathrm{I}} \times \frac{2 \mathrm{R}}{\mathrm{R}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{1}$ $\mathrm{~B}_{1}: \mathrm{B}_{2}=1: 1$
BCECE-2014
Moving Charges & Magnetism
153284
A horizontal overhead power line carries acurrent of $90 \mathrm{~A}$ in east to west direction. Magnitude of magnetic field due to the current $1.5 \mathrm{~m}$ below the line is-
1 $1.2 \mathrm{~T}$
2 $1.2 \times 10^{-10} \mathrm{~T}$
3 $0 \mathrm{~T}$
4 $1.2 \times 10^{-5} \mathrm{~T}$
Explanation:
D Given that, $\mathrm{I}=90 \mathrm{~A}, \mathrm{~d}=1.5 \mathrm{~m}$ Magnitude of magnetic field at a distance $d$ due to straight current carrying conductor is given by- $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}$ $\mathrm{B}=\frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$ $\mathrm{~B}=2 \times 10^{-7} \times 60$ $\mathrm{~B}=120 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}=1.2 \times 10^{-5} \mathrm{~T}$
BCECE-2014
Moving Charges & Magnetism
153285
Which of the following graphs shows the variation of magnetic field $B$, with distance from a long current carrying conductor?
1 a
2 b
3 c
4 d
Explanation:
C Magnetic field due to long current carrying conductor at a distance $r$ is- $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ So, the graph between the magnetic field and the distance $r$ is-
COMEDK 2015
Moving Charges & Magnetism
153286
The distance between the poles of a horse shoe magnet is $0.1 \mathrm{~m}$ and its pole strength is $0.01 \mathrm{~A}$ $m$. The induction of magnetic field at a point mid way between the poles will be-
1 Zero
2 $2 \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-6} \mathrm{~T}$
4 $8 \times 10^{-7} \mathrm{~T}$
Explanation:
D Given, $\mathrm{d}=0.1 \mathrm{~m} \quad \mathrm{r}=\frac{0.1}{2}=0.05 \mathrm{~m}$ Pole strength $=\mathrm{m}=0.01 \quad$ A-m Net magnetic field at midpoint $\mathrm{O}$, $\mathrm{B}_{\text {net }}=\mathrm{B}_{\mathrm{N}}+\mathrm{B}_{\mathrm{S}}$ Where, $\mathrm{B}_{\mathrm{N}}=$ Magnitude of magnetic field due to $\mathrm{N}$ Pole $\mathrm{B}_{\mathrm{S}}=$ Magnitude of magnetic field due to $\mathrm{S}$ pole. $\mathrm{B}_{\mathrm{S}}=\mathrm{B}_{\mathrm{N}}=\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}}$ $\mathrm{~B}=\frac{4 \pi \times 10^{-7} \times 0.01}{4 \pi\left(\frac{0.1}{2}\right)^{2}}=4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{\text {net }}=2 \mathrm{~B}=2 \times 4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{\text {net }}=8 \times 10^{-7} \mathrm{~T}$
