153156
A horizontal overhead power line at a height of $5 \mathrm{~m}$ from the ground and carries a current of 150 A from East to West. The magnetic field directly below it on ground is
1 $6 \times 10^{-6} \mathrm{~T}$ towards South
2 $6 \times 10^{-6} \mathrm{~T}$ towards West
3 $7 \times 10^{-6} \mathrm{~T}$ towards East
4 $8 \times 10^{-7} \mathrm{~T}$ towards North
Explanation:
A From the right hand rule of palm direction of magnetic field will be towards in south direction and for finding the value of it. $\text { Direction }=\text { Southward }$ $\mathrm{B}=\frac{\mu_{0} 2 \mathrm{I}}{4 \pi \mathrm{r}}=\frac{10^{-7} \times 2 \times 150}{5}$ $\mathrm{~B}=6 \times 10^{-6} \mathrm{~T}$ Towards south
AP EAMCET (18.09.2020) Shift-I
Moving Charges & Magnetism
153157
A proton moving with a velocity $2.5 \times 10^{7} \mathrm{~m} / \mathrm{s}$ enters a magnetic field of intensity $2.5 \mathrm{~T}$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is
153159
10 A current is passing through a very long wire of radius $5 \mathrm{~cm}$. Then magnetic field at a distance of $2 \mathrm{~cm}$ inside from its curved surface is $\times 10^{-5} \mathrm{~T}$.
153161
Two long parallel straight wires are carrying current $i$ each but in opposite direction, If the distance between the wires is $2 r$, then the magnitude of magnetic field midway between them will be
1 zero
2 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$
3 $\frac{\mu_{0} \mathrm{i}}{\pi \mathrm{r}}$
4 $\frac{4 \mu_{0} \mathrm{i}}{\pi \mathrm{r}}$
Explanation:
C When currents flows in possite direction, then the magnetic field produced are additive in nature, so $\mathrm{B}_{\text {net }}=2 \times \frac{\mu_{0} \mathrm{I}}{2 \pi \times\left(\frac{2 \mathrm{R}}{2}\right)}$ $\mathrm{B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{\pi \mathrm{R}}$
153156
A horizontal overhead power line at a height of $5 \mathrm{~m}$ from the ground and carries a current of 150 A from East to West. The magnetic field directly below it on ground is
1 $6 \times 10^{-6} \mathrm{~T}$ towards South
2 $6 \times 10^{-6} \mathrm{~T}$ towards West
3 $7 \times 10^{-6} \mathrm{~T}$ towards East
4 $8 \times 10^{-7} \mathrm{~T}$ towards North
Explanation:
A From the right hand rule of palm direction of magnetic field will be towards in south direction and for finding the value of it. $\text { Direction }=\text { Southward }$ $\mathrm{B}=\frac{\mu_{0} 2 \mathrm{I}}{4 \pi \mathrm{r}}=\frac{10^{-7} \times 2 \times 150}{5}$ $\mathrm{~B}=6 \times 10^{-6} \mathrm{~T}$ Towards south
AP EAMCET (18.09.2020) Shift-I
Moving Charges & Magnetism
153157
A proton moving with a velocity $2.5 \times 10^{7} \mathrm{~m} / \mathrm{s}$ enters a magnetic field of intensity $2.5 \mathrm{~T}$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is
153159
10 A current is passing through a very long wire of radius $5 \mathrm{~cm}$. Then magnetic field at a distance of $2 \mathrm{~cm}$ inside from its curved surface is $\times 10^{-5} \mathrm{~T}$.
153161
Two long parallel straight wires are carrying current $i$ each but in opposite direction, If the distance between the wires is $2 r$, then the magnitude of magnetic field midway between them will be
1 zero
2 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$
3 $\frac{\mu_{0} \mathrm{i}}{\pi \mathrm{r}}$
4 $\frac{4 \mu_{0} \mathrm{i}}{\pi \mathrm{r}}$
Explanation:
C When currents flows in possite direction, then the magnetic field produced are additive in nature, so $\mathrm{B}_{\text {net }}=2 \times \frac{\mu_{0} \mathrm{I}}{2 \pi \times\left(\frac{2 \mathrm{R}}{2}\right)}$ $\mathrm{B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{\pi \mathrm{R}}$
153156
A horizontal overhead power line at a height of $5 \mathrm{~m}$ from the ground and carries a current of 150 A from East to West. The magnetic field directly below it on ground is
1 $6 \times 10^{-6} \mathrm{~T}$ towards South
2 $6 \times 10^{-6} \mathrm{~T}$ towards West
3 $7 \times 10^{-6} \mathrm{~T}$ towards East
4 $8 \times 10^{-7} \mathrm{~T}$ towards North
Explanation:
A From the right hand rule of palm direction of magnetic field will be towards in south direction and for finding the value of it. $\text { Direction }=\text { Southward }$ $\mathrm{B}=\frac{\mu_{0} 2 \mathrm{I}}{4 \pi \mathrm{r}}=\frac{10^{-7} \times 2 \times 150}{5}$ $\mathrm{~B}=6 \times 10^{-6} \mathrm{~T}$ Towards south
AP EAMCET (18.09.2020) Shift-I
Moving Charges & Magnetism
153157
A proton moving with a velocity $2.5 \times 10^{7} \mathrm{~m} / \mathrm{s}$ enters a magnetic field of intensity $2.5 \mathrm{~T}$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is
153159
10 A current is passing through a very long wire of radius $5 \mathrm{~cm}$. Then magnetic field at a distance of $2 \mathrm{~cm}$ inside from its curved surface is $\times 10^{-5} \mathrm{~T}$.
153161
Two long parallel straight wires are carrying current $i$ each but in opposite direction, If the distance between the wires is $2 r$, then the magnitude of magnetic field midway between them will be
1 zero
2 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$
3 $\frac{\mu_{0} \mathrm{i}}{\pi \mathrm{r}}$
4 $\frac{4 \mu_{0} \mathrm{i}}{\pi \mathrm{r}}$
Explanation:
C When currents flows in possite direction, then the magnetic field produced are additive in nature, so $\mathrm{B}_{\text {net }}=2 \times \frac{\mu_{0} \mathrm{I}}{2 \pi \times\left(\frac{2 \mathrm{R}}{2}\right)}$ $\mathrm{B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{\pi \mathrm{R}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153156
A horizontal overhead power line at a height of $5 \mathrm{~m}$ from the ground and carries a current of 150 A from East to West. The magnetic field directly below it on ground is
1 $6 \times 10^{-6} \mathrm{~T}$ towards South
2 $6 \times 10^{-6} \mathrm{~T}$ towards West
3 $7 \times 10^{-6} \mathrm{~T}$ towards East
4 $8 \times 10^{-7} \mathrm{~T}$ towards North
Explanation:
A From the right hand rule of palm direction of magnetic field will be towards in south direction and for finding the value of it. $\text { Direction }=\text { Southward }$ $\mathrm{B}=\frac{\mu_{0} 2 \mathrm{I}}{4 \pi \mathrm{r}}=\frac{10^{-7} \times 2 \times 150}{5}$ $\mathrm{~B}=6 \times 10^{-6} \mathrm{~T}$ Towards south
AP EAMCET (18.09.2020) Shift-I
Moving Charges & Magnetism
153157
A proton moving with a velocity $2.5 \times 10^{7} \mathrm{~m} / \mathrm{s}$ enters a magnetic field of intensity $2.5 \mathrm{~T}$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is
153159
10 A current is passing through a very long wire of radius $5 \mathrm{~cm}$. Then magnetic field at a distance of $2 \mathrm{~cm}$ inside from its curved surface is $\times 10^{-5} \mathrm{~T}$.
153161
Two long parallel straight wires are carrying current $i$ each but in opposite direction, If the distance between the wires is $2 r$, then the magnitude of magnetic field midway between them will be
1 zero
2 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$
3 $\frac{\mu_{0} \mathrm{i}}{\pi \mathrm{r}}$
4 $\frac{4 \mu_{0} \mathrm{i}}{\pi \mathrm{r}}$
Explanation:
C When currents flows in possite direction, then the magnetic field produced are additive in nature, so $\mathrm{B}_{\text {net }}=2 \times \frac{\mu_{0} \mathrm{I}}{2 \pi \times\left(\frac{2 \mathrm{R}}{2}\right)}$ $\mathrm{B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{\pi \mathrm{R}}$
