362549
A current loop, having two circular arc is joined by two radial lines (as shown in fig.) and carries a current of \(10A\). The magnetic field at point \(0\) is \(N \times {10^{ - 5}}\;T\). Then find \(N.\)
362550
Two similar coils of radius \(R\) are lying concentrically with their planes at right angles to each other. The currents flowing in them are \(I\) and 2\(I\), respectively. The resultant magnetic field induction at the centre will be
1 \(\dfrac{\mu_{0} I}{R}\)
2 \(\dfrac{\mu_{0} I}{2 R}\)
3 \(\dfrac{\sqrt{5} \mu_{0} I}{2 R}\)
4 \(\dfrac{3 \mu_{0} I}{2 R}\)
Explanation:
The magnetic field (2) at the centre of circular current carrying coil of radius \(\mathrm{R}\) and current \(\mathrm{I}\) is \(B=\dfrac{\mu_{0} I}{2 R}\) Similarly, if current is \(2 \mathrm{I}\), then Magnetic field \(=\dfrac{\mu_{0} 2 I}{2 R}=2 B\) So, resultant magnetic field \(\begin{aligned}& =\sqrt{B^{2}+(2 B)^{2}}=\sqrt{5 B^{2}}=\sqrt{5} B \\& =\dfrac{\mu_{0} I \sqrt{5}}{2 R}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362551
A conducting wire carrying current \(I\) is arranged as shown. The magnetic field at \(O\)
362552
Magnetic field at the centre of a circular loop of area \(A\) is \(B\). The magnetic moment of the loop will be
1 \(\dfrac{B A^{2}}{\mu_{0} \pi}\)
2 \(\dfrac{B A^{3 / 2}}{\mu_{0} \pi}\)
3 \(\dfrac{B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
4 \(\dfrac{2 B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
Explanation:
Magnetic field at the centre of circular loop, \(B=\dfrac{\mu_{0}}{4 \pi} \dfrac{2 \pi I}{r}=\dfrac{\mu_{0} I}{2 r} \Rightarrow I=\dfrac{2 B r}{\mu_{0}}\) Also, \(A=\pi r^{2}\) or \(r=(A / \pi)^{1 / 2}\) Magnetic moment, \(M=I A=\dfrac{2 B r}{\mu_{0}} A\) \(=\dfrac{2 B A}{\mu_{0}} \times\left(\dfrac{A}{\pi}\right)^{1 / 2}=\dfrac{2 B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
362549
A current loop, having two circular arc is joined by two radial lines (as shown in fig.) and carries a current of \(10A\). The magnetic field at point \(0\) is \(N \times {10^{ - 5}}\;T\). Then find \(N.\)
362550
Two similar coils of radius \(R\) are lying concentrically with their planes at right angles to each other. The currents flowing in them are \(I\) and 2\(I\), respectively. The resultant magnetic field induction at the centre will be
1 \(\dfrac{\mu_{0} I}{R}\)
2 \(\dfrac{\mu_{0} I}{2 R}\)
3 \(\dfrac{\sqrt{5} \mu_{0} I}{2 R}\)
4 \(\dfrac{3 \mu_{0} I}{2 R}\)
Explanation:
The magnetic field (2) at the centre of circular current carrying coil of radius \(\mathrm{R}\) and current \(\mathrm{I}\) is \(B=\dfrac{\mu_{0} I}{2 R}\) Similarly, if current is \(2 \mathrm{I}\), then Magnetic field \(=\dfrac{\mu_{0} 2 I}{2 R}=2 B\) So, resultant magnetic field \(\begin{aligned}& =\sqrt{B^{2}+(2 B)^{2}}=\sqrt{5 B^{2}}=\sqrt{5} B \\& =\dfrac{\mu_{0} I \sqrt{5}}{2 R}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362551
A conducting wire carrying current \(I\) is arranged as shown. The magnetic field at \(O\)
362552
Magnetic field at the centre of a circular loop of area \(A\) is \(B\). The magnetic moment of the loop will be
1 \(\dfrac{B A^{2}}{\mu_{0} \pi}\)
2 \(\dfrac{B A^{3 / 2}}{\mu_{0} \pi}\)
3 \(\dfrac{B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
4 \(\dfrac{2 B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
Explanation:
Magnetic field at the centre of circular loop, \(B=\dfrac{\mu_{0}}{4 \pi} \dfrac{2 \pi I}{r}=\dfrac{\mu_{0} I}{2 r} \Rightarrow I=\dfrac{2 B r}{\mu_{0}}\) Also, \(A=\pi r^{2}\) or \(r=(A / \pi)^{1 / 2}\) Magnetic moment, \(M=I A=\dfrac{2 B r}{\mu_{0}} A\) \(=\dfrac{2 B A}{\mu_{0}} \times\left(\dfrac{A}{\pi}\right)^{1 / 2}=\dfrac{2 B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
362549
A current loop, having two circular arc is joined by two radial lines (as shown in fig.) and carries a current of \(10A\). The magnetic field at point \(0\) is \(N \times {10^{ - 5}}\;T\). Then find \(N.\)
362550
Two similar coils of radius \(R\) are lying concentrically with their planes at right angles to each other. The currents flowing in them are \(I\) and 2\(I\), respectively. The resultant magnetic field induction at the centre will be
1 \(\dfrac{\mu_{0} I}{R}\)
2 \(\dfrac{\mu_{0} I}{2 R}\)
3 \(\dfrac{\sqrt{5} \mu_{0} I}{2 R}\)
4 \(\dfrac{3 \mu_{0} I}{2 R}\)
Explanation:
The magnetic field (2) at the centre of circular current carrying coil of radius \(\mathrm{R}\) and current \(\mathrm{I}\) is \(B=\dfrac{\mu_{0} I}{2 R}\) Similarly, if current is \(2 \mathrm{I}\), then Magnetic field \(=\dfrac{\mu_{0} 2 I}{2 R}=2 B\) So, resultant magnetic field \(\begin{aligned}& =\sqrt{B^{2}+(2 B)^{2}}=\sqrt{5 B^{2}}=\sqrt{5} B \\& =\dfrac{\mu_{0} I \sqrt{5}}{2 R}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362551
A conducting wire carrying current \(I\) is arranged as shown. The magnetic field at \(O\)
362552
Magnetic field at the centre of a circular loop of area \(A\) is \(B\). The magnetic moment of the loop will be
1 \(\dfrac{B A^{2}}{\mu_{0} \pi}\)
2 \(\dfrac{B A^{3 / 2}}{\mu_{0} \pi}\)
3 \(\dfrac{B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
4 \(\dfrac{2 B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
Explanation:
Magnetic field at the centre of circular loop, \(B=\dfrac{\mu_{0}}{4 \pi} \dfrac{2 \pi I}{r}=\dfrac{\mu_{0} I}{2 r} \Rightarrow I=\dfrac{2 B r}{\mu_{0}}\) Also, \(A=\pi r^{2}\) or \(r=(A / \pi)^{1 / 2}\) Magnetic moment, \(M=I A=\dfrac{2 B r}{\mu_{0}} A\) \(=\dfrac{2 B A}{\mu_{0}} \times\left(\dfrac{A}{\pi}\right)^{1 / 2}=\dfrac{2 B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII04:MOVING CHARGES AND MAGNETISM
362549
A current loop, having two circular arc is joined by two radial lines (as shown in fig.) and carries a current of \(10A\). The magnetic field at point \(0\) is \(N \times {10^{ - 5}}\;T\). Then find \(N.\)
362550
Two similar coils of radius \(R\) are lying concentrically with their planes at right angles to each other. The currents flowing in them are \(I\) and 2\(I\), respectively. The resultant magnetic field induction at the centre will be
1 \(\dfrac{\mu_{0} I}{R}\)
2 \(\dfrac{\mu_{0} I}{2 R}\)
3 \(\dfrac{\sqrt{5} \mu_{0} I}{2 R}\)
4 \(\dfrac{3 \mu_{0} I}{2 R}\)
Explanation:
The magnetic field (2) at the centre of circular current carrying coil of radius \(\mathrm{R}\) and current \(\mathrm{I}\) is \(B=\dfrac{\mu_{0} I}{2 R}\) Similarly, if current is \(2 \mathrm{I}\), then Magnetic field \(=\dfrac{\mu_{0} 2 I}{2 R}=2 B\) So, resultant magnetic field \(\begin{aligned}& =\sqrt{B^{2}+(2 B)^{2}}=\sqrt{5 B^{2}}=\sqrt{5} B \\& =\dfrac{\mu_{0} I \sqrt{5}}{2 R}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362551
A conducting wire carrying current \(I\) is arranged as shown. The magnetic field at \(O\)
362552
Magnetic field at the centre of a circular loop of area \(A\) is \(B\). The magnetic moment of the loop will be
1 \(\dfrac{B A^{2}}{\mu_{0} \pi}\)
2 \(\dfrac{B A^{3 / 2}}{\mu_{0} \pi}\)
3 \(\dfrac{B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
4 \(\dfrac{2 B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
Explanation:
Magnetic field at the centre of circular loop, \(B=\dfrac{\mu_{0}}{4 \pi} \dfrac{2 \pi I}{r}=\dfrac{\mu_{0} I}{2 r} \Rightarrow I=\dfrac{2 B r}{\mu_{0}}\) Also, \(A=\pi r^{2}\) or \(r=(A / \pi)^{1 / 2}\) Magnetic moment, \(M=I A=\dfrac{2 B r}{\mu_{0}} A\) \(=\dfrac{2 B A}{\mu_{0}} \times\left(\dfrac{A}{\pi}\right)^{1 / 2}=\dfrac{2 B A^{3 / 2}}{\mu_{0} \pi^{1 / 2}}\)
