NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII01:ELECTRIC CHARGES AND FIELDS
358282
A uniform electrical field \(a\widehat i + b\widehat j\) intersects a surface of area \(A\). What is the flux through this area if the surface lies in \(XZ\) plane?
1 \(Ab\)
2 \(Aa\)
3 \(\frac{{A\left( {ab} \right)}}{{a + b}}\)
4 \(A\sqrt {{a^2} + {b^2}} \)
Explanation:
The direction of the field is Given that \(\overline {\rm{E}} = a\hat i + b\hat j\,\& \overline A = A\hat j\) \(\phi = \overline E .\overline A = Ab\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358283
A loop of diameter \(d\) is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be \(\phi \). What is the electric field strength?
1 \(\frac{{4\phi }}{{\pi {d^2}}}\)
2 \(\frac{{2\phi }}{{\pi {d^2}}}\)
3 \(\frac{\phi }{{\pi {d^2}}}\)
4 \(\frac{{\pi \phi {d^2}}}{4}\)
Explanation:
\(\phi = EA\cos {0^ \circ } = E \times \frac{{\pi {d^2}}}{4},\therefore E = \frac{{4\phi }}{{\pi {d^2}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358284
An electric field is given by, \((6 \hat{i}+5 \hat{j}+3 \hat{k}) N / C\). The electric flux through a surface area \(30\,\hat i\;\,{m^2}\) lying in \(Y\,Z - \) plane (in \(SI\) unit) is
1 150
2 60
3 180
4 90
Explanation:
Given : \(\vec E = 6\,\hat i + 5\,\hat j + 3\,\hat k\,N/C,\vec A = 30\,\hat i\,{m^2}\) Electric flux, \(\phi = \vec E\,.\,\,\vec A\) \(\phi = (6\,\hat i + 5\hat j + 3\,\hat k) \cdot (30\,\hat i) = 180\,N{m^2}/C\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358285
A uniform electric field \(E = 2 \times {10^3}N{C^{ - 1}}\) is acting along the positive \(x\)-axis. The flux of this field through a square of \(10\,cm\) side whose plane is parallel to the \(yz\) plane is
358282
A uniform electrical field \(a\widehat i + b\widehat j\) intersects a surface of area \(A\). What is the flux through this area if the surface lies in \(XZ\) plane?
1 \(Ab\)
2 \(Aa\)
3 \(\frac{{A\left( {ab} \right)}}{{a + b}}\)
4 \(A\sqrt {{a^2} + {b^2}} \)
Explanation:
The direction of the field is Given that \(\overline {\rm{E}} = a\hat i + b\hat j\,\& \overline A = A\hat j\) \(\phi = \overline E .\overline A = Ab\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358283
A loop of diameter \(d\) is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be \(\phi \). What is the electric field strength?
1 \(\frac{{4\phi }}{{\pi {d^2}}}\)
2 \(\frac{{2\phi }}{{\pi {d^2}}}\)
3 \(\frac{\phi }{{\pi {d^2}}}\)
4 \(\frac{{\pi \phi {d^2}}}{4}\)
Explanation:
\(\phi = EA\cos {0^ \circ } = E \times \frac{{\pi {d^2}}}{4},\therefore E = \frac{{4\phi }}{{\pi {d^2}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358284
An electric field is given by, \((6 \hat{i}+5 \hat{j}+3 \hat{k}) N / C\). The electric flux through a surface area \(30\,\hat i\;\,{m^2}\) lying in \(Y\,Z - \) plane (in \(SI\) unit) is
1 150
2 60
3 180
4 90
Explanation:
Given : \(\vec E = 6\,\hat i + 5\,\hat j + 3\,\hat k\,N/C,\vec A = 30\,\hat i\,{m^2}\) Electric flux, \(\phi = \vec E\,.\,\,\vec A\) \(\phi = (6\,\hat i + 5\hat j + 3\,\hat k) \cdot (30\,\hat i) = 180\,N{m^2}/C\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358285
A uniform electric field \(E = 2 \times {10^3}N{C^{ - 1}}\) is acting along the positive \(x\)-axis. The flux of this field through a square of \(10\,cm\) side whose plane is parallel to the \(yz\) plane is
358282
A uniform electrical field \(a\widehat i + b\widehat j\) intersects a surface of area \(A\). What is the flux through this area if the surface lies in \(XZ\) plane?
1 \(Ab\)
2 \(Aa\)
3 \(\frac{{A\left( {ab} \right)}}{{a + b}}\)
4 \(A\sqrt {{a^2} + {b^2}} \)
Explanation:
The direction of the field is Given that \(\overline {\rm{E}} = a\hat i + b\hat j\,\& \overline A = A\hat j\) \(\phi = \overline E .\overline A = Ab\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358283
A loop of diameter \(d\) is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be \(\phi \). What is the electric field strength?
1 \(\frac{{4\phi }}{{\pi {d^2}}}\)
2 \(\frac{{2\phi }}{{\pi {d^2}}}\)
3 \(\frac{\phi }{{\pi {d^2}}}\)
4 \(\frac{{\pi \phi {d^2}}}{4}\)
Explanation:
\(\phi = EA\cos {0^ \circ } = E \times \frac{{\pi {d^2}}}{4},\therefore E = \frac{{4\phi }}{{\pi {d^2}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358284
An electric field is given by, \((6 \hat{i}+5 \hat{j}+3 \hat{k}) N / C\). The electric flux through a surface area \(30\,\hat i\;\,{m^2}\) lying in \(Y\,Z - \) plane (in \(SI\) unit) is
1 150
2 60
3 180
4 90
Explanation:
Given : \(\vec E = 6\,\hat i + 5\,\hat j + 3\,\hat k\,N/C,\vec A = 30\,\hat i\,{m^2}\) Electric flux, \(\phi = \vec E\,.\,\,\vec A\) \(\phi = (6\,\hat i + 5\hat j + 3\,\hat k) \cdot (30\,\hat i) = 180\,N{m^2}/C\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358285
A uniform electric field \(E = 2 \times {10^3}N{C^{ - 1}}\) is acting along the positive \(x\)-axis. The flux of this field through a square of \(10\,cm\) side whose plane is parallel to the \(yz\) plane is
358282
A uniform electrical field \(a\widehat i + b\widehat j\) intersects a surface of area \(A\). What is the flux through this area if the surface lies in \(XZ\) plane?
1 \(Ab\)
2 \(Aa\)
3 \(\frac{{A\left( {ab} \right)}}{{a + b}}\)
4 \(A\sqrt {{a^2} + {b^2}} \)
Explanation:
The direction of the field is Given that \(\overline {\rm{E}} = a\hat i + b\hat j\,\& \overline A = A\hat j\) \(\phi = \overline E .\overline A = Ab\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358283
A loop of diameter \(d\) is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be \(\phi \). What is the electric field strength?
1 \(\frac{{4\phi }}{{\pi {d^2}}}\)
2 \(\frac{{2\phi }}{{\pi {d^2}}}\)
3 \(\frac{\phi }{{\pi {d^2}}}\)
4 \(\frac{{\pi \phi {d^2}}}{4}\)
Explanation:
\(\phi = EA\cos {0^ \circ } = E \times \frac{{\pi {d^2}}}{4},\therefore E = \frac{{4\phi }}{{\pi {d^2}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358284
An electric field is given by, \((6 \hat{i}+5 \hat{j}+3 \hat{k}) N / C\). The electric flux through a surface area \(30\,\hat i\;\,{m^2}\) lying in \(Y\,Z - \) plane (in \(SI\) unit) is
1 150
2 60
3 180
4 90
Explanation:
Given : \(\vec E = 6\,\hat i + 5\,\hat j + 3\,\hat k\,N/C,\vec A = 30\,\hat i\,{m^2}\) Electric flux, \(\phi = \vec E\,.\,\,\vec A\) \(\phi = (6\,\hat i + 5\hat j + 3\,\hat k) \cdot (30\,\hat i) = 180\,N{m^2}/C\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358285
A uniform electric field \(E = 2 \times {10^3}N{C^{ - 1}}\) is acting along the positive \(x\)-axis. The flux of this field through a square of \(10\,cm\) side whose plane is parallel to the \(yz\) plane is
