362587
A circular wire of radius \(R\) is made with uniform material. A current enters at \(A\) and leaves from \(B\). The magnetic field at the centre \(C\) is
1 \(\dfrac{\mu_{0} I}{2 R}\)
2 \(\dfrac{\mu_{0} I}{4 R}\)
3 \(\dfrac{\mu_{0} I}{R}\)
4 Zero
Explanation:
\(\overrightarrow {{B_1}} + \overrightarrow {{B_2}} = 0\) \({B_1} \to \) due to left part \({B_2} \to \) due to right part \(B_{1}, B_{2}\) at centre are numerically equal \(\&\) opposite to each other. (i,e. one is into the paper and the other is towards the reader)
PHXII04:MOVING CHARGES AND MAGNETISM
362588
A wire bent as shown in the figure carries a current \(I\). The magnetic field at \(P\) is
1 \(\dfrac{3 \mu_{0} I}{2 R}\)
2 \(\dfrac{\mu_{0} I}{4 R}\)
3 \(\dfrac{7 \mu_{0} I}{8 R}\)
4 \(\dfrac{\mu_{0} I}{8 R}\)
Explanation:
Let us divide the given conductor into 3 parts. \(B\) due to straight parts \(\left(B_{1}, B_{2}\right)\) is zero and \(B\) due arc is \(\begin{gathered}B_{2}=\left(\dfrac{\mu_{o} I}{2 R} \dfrac{\pi / 2}{2 \pi}\right) \\B=B_{1}+B_{2}+B_{3}=0+\dfrac{\mu_{0} I}{2 R}\left(\dfrac{\pi / 2}{2 \pi}\right)+0=\dfrac{\mu_{0} I}{8 R}\end{gathered}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362589
In the loop shown, the magnetic induction at the point \('O'\) is
Since the point \(O\) lies on the lines of the straight wires, they contribute no magnetic induction at \(O\). The magnetic induction at \(\mathrm{O}\) due to arcs only. The magnetic induction at \(O\) due to are of radius \(R_{1}\) is \({B_1} = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{{{R_1}}}\frac{\pi }{2} = \frac{{{\mu _0}I}}{{8{R_1}}} \oplus \) The magnetic field at \(O\) due to arc of radius\({R_2}\) is \(B_{2}=\dfrac{\mu_{0}}{4 \pi} \dfrac{I}{R_{1}} \dfrac{\pi}{2}=\dfrac{\mu_{0} I}{8 R_{2}} \otimes\) The net magnetic field at \(O\) is \({B_1} + {B_2}\) \(B = \frac{{{\mu _0}I}}{{8{R_1}}} + \frac{{{\mu _0}I}}{{8{R_2}}} = \frac{{{\mu _0}I}}{8}\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right) = \frac{{{\mu _0}I}}{8}\left( {\frac{{{R_2} + {R_1}}}{{{R_1}{R_2}}}} \right)\)
KCET - 2012
PHXII04:MOVING CHARGES AND MAGNETISM
362590
A long wire having a semicircular loop of radius \(r\) carries a current \(i\) as shown in figure. The magnetic induction at the centre \(O\) due to entire wire is
1 \(\dfrac{\mu_{0} i}{4 r}\)
2 \(\frac{{{\mu _0}{i^2}}}{{4{r^2}}}\)
3 \(\frac{{{\mu _0}i}}{{4r}}\)
4 None of these
Explanation:
According to Biot-Savart's law, the magnetic induction at a point to a current carrying element \(i \delta l\) is given by \(\delta B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{i\delta 1 \times r}}{{{r^3}}}\) or \(B=\dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l \sin \theta}{r^{2}}\) Directed normal to plane containing \(\delta 1\) and \(r,\theta \) being angle between \(\delta 1\) and \(r\). Field due to semicircular arc Now, angle between a current element \(\delta 1\) of semicircular arc and the radius vector of the element to point \(c\) is \(\pi / 2\). Therefore, the magnitude of magnetic induction \(B\) at \(O\) due to this element is \(\delta B=\dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l \sin \pi / 2}{r^{2}}=\dfrac{\mu_{0} i \delta l}{4 \pi r^{2}}\) Hence, magnetic induction due to whole semicircular loop is \(\begin{aligned}B & =\Sigma \delta B=\Sigma \dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l}{r^{2}} \\& =\dfrac{\mu_{0}}{4 \pi} \dfrac{i}{r^{2}} \Sigma \delta l \\& =\dfrac{\mu_{0} i}{4 \pi r^{2}}(\pi r)=\dfrac{\mu_{0} i}{4 r}\end{aligned}\) The magnetic field due to \(a b\) and \(d e\) is zero, because \(\theta=0^{\circ}\) or \(180^{\circ}\). So, net magnetic field is \(B=\dfrac{\mu_{0} i}{4 r}\)
362587
A circular wire of radius \(R\) is made with uniform material. A current enters at \(A\) and leaves from \(B\). The magnetic field at the centre \(C\) is
1 \(\dfrac{\mu_{0} I}{2 R}\)
2 \(\dfrac{\mu_{0} I}{4 R}\)
3 \(\dfrac{\mu_{0} I}{R}\)
4 Zero
Explanation:
\(\overrightarrow {{B_1}} + \overrightarrow {{B_2}} = 0\) \({B_1} \to \) due to left part \({B_2} \to \) due to right part \(B_{1}, B_{2}\) at centre are numerically equal \(\&\) opposite to each other. (i,e. one is into the paper and the other is towards the reader)
PHXII04:MOVING CHARGES AND MAGNETISM
362588
A wire bent as shown in the figure carries a current \(I\). The magnetic field at \(P\) is
1 \(\dfrac{3 \mu_{0} I}{2 R}\)
2 \(\dfrac{\mu_{0} I}{4 R}\)
3 \(\dfrac{7 \mu_{0} I}{8 R}\)
4 \(\dfrac{\mu_{0} I}{8 R}\)
Explanation:
Let us divide the given conductor into 3 parts. \(B\) due to straight parts \(\left(B_{1}, B_{2}\right)\) is zero and \(B\) due arc is \(\begin{gathered}B_{2}=\left(\dfrac{\mu_{o} I}{2 R} \dfrac{\pi / 2}{2 \pi}\right) \\B=B_{1}+B_{2}+B_{3}=0+\dfrac{\mu_{0} I}{2 R}\left(\dfrac{\pi / 2}{2 \pi}\right)+0=\dfrac{\mu_{0} I}{8 R}\end{gathered}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362589
In the loop shown, the magnetic induction at the point \('O'\) is
Since the point \(O\) lies on the lines of the straight wires, they contribute no magnetic induction at \(O\). The magnetic induction at \(\mathrm{O}\) due to arcs only. The magnetic induction at \(O\) due to are of radius \(R_{1}\) is \({B_1} = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{{{R_1}}}\frac{\pi }{2} = \frac{{{\mu _0}I}}{{8{R_1}}} \oplus \) The magnetic field at \(O\) due to arc of radius\({R_2}\) is \(B_{2}=\dfrac{\mu_{0}}{4 \pi} \dfrac{I}{R_{1}} \dfrac{\pi}{2}=\dfrac{\mu_{0} I}{8 R_{2}} \otimes\) The net magnetic field at \(O\) is \({B_1} + {B_2}\) \(B = \frac{{{\mu _0}I}}{{8{R_1}}} + \frac{{{\mu _0}I}}{{8{R_2}}} = \frac{{{\mu _0}I}}{8}\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right) = \frac{{{\mu _0}I}}{8}\left( {\frac{{{R_2} + {R_1}}}{{{R_1}{R_2}}}} \right)\)
KCET - 2012
PHXII04:MOVING CHARGES AND MAGNETISM
362590
A long wire having a semicircular loop of radius \(r\) carries a current \(i\) as shown in figure. The magnetic induction at the centre \(O\) due to entire wire is
1 \(\dfrac{\mu_{0} i}{4 r}\)
2 \(\frac{{{\mu _0}{i^2}}}{{4{r^2}}}\)
3 \(\frac{{{\mu _0}i}}{{4r}}\)
4 None of these
Explanation:
According to Biot-Savart's law, the magnetic induction at a point to a current carrying element \(i \delta l\) is given by \(\delta B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{i\delta 1 \times r}}{{{r^3}}}\) or \(B=\dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l \sin \theta}{r^{2}}\) Directed normal to plane containing \(\delta 1\) and \(r,\theta \) being angle between \(\delta 1\) and \(r\). Field due to semicircular arc Now, angle between a current element \(\delta 1\) of semicircular arc and the radius vector of the element to point \(c\) is \(\pi / 2\). Therefore, the magnitude of magnetic induction \(B\) at \(O\) due to this element is \(\delta B=\dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l \sin \pi / 2}{r^{2}}=\dfrac{\mu_{0} i \delta l}{4 \pi r^{2}}\) Hence, magnetic induction due to whole semicircular loop is \(\begin{aligned}B & =\Sigma \delta B=\Sigma \dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l}{r^{2}} \\& =\dfrac{\mu_{0}}{4 \pi} \dfrac{i}{r^{2}} \Sigma \delta l \\& =\dfrac{\mu_{0} i}{4 \pi r^{2}}(\pi r)=\dfrac{\mu_{0} i}{4 r}\end{aligned}\) The magnetic field due to \(a b\) and \(d e\) is zero, because \(\theta=0^{\circ}\) or \(180^{\circ}\). So, net magnetic field is \(B=\dfrac{\mu_{0} i}{4 r}\)
362587
A circular wire of radius \(R\) is made with uniform material. A current enters at \(A\) and leaves from \(B\). The magnetic field at the centre \(C\) is
1 \(\dfrac{\mu_{0} I}{2 R}\)
2 \(\dfrac{\mu_{0} I}{4 R}\)
3 \(\dfrac{\mu_{0} I}{R}\)
4 Zero
Explanation:
\(\overrightarrow {{B_1}} + \overrightarrow {{B_2}} = 0\) \({B_1} \to \) due to left part \({B_2} \to \) due to right part \(B_{1}, B_{2}\) at centre are numerically equal \(\&\) opposite to each other. (i,e. one is into the paper and the other is towards the reader)
PHXII04:MOVING CHARGES AND MAGNETISM
362588
A wire bent as shown in the figure carries a current \(I\). The magnetic field at \(P\) is
1 \(\dfrac{3 \mu_{0} I}{2 R}\)
2 \(\dfrac{\mu_{0} I}{4 R}\)
3 \(\dfrac{7 \mu_{0} I}{8 R}\)
4 \(\dfrac{\mu_{0} I}{8 R}\)
Explanation:
Let us divide the given conductor into 3 parts. \(B\) due to straight parts \(\left(B_{1}, B_{2}\right)\) is zero and \(B\) due arc is \(\begin{gathered}B_{2}=\left(\dfrac{\mu_{o} I}{2 R} \dfrac{\pi / 2}{2 \pi}\right) \\B=B_{1}+B_{2}+B_{3}=0+\dfrac{\mu_{0} I}{2 R}\left(\dfrac{\pi / 2}{2 \pi}\right)+0=\dfrac{\mu_{0} I}{8 R}\end{gathered}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362589
In the loop shown, the magnetic induction at the point \('O'\) is
Since the point \(O\) lies on the lines of the straight wires, they contribute no magnetic induction at \(O\). The magnetic induction at \(\mathrm{O}\) due to arcs only. The magnetic induction at \(O\) due to are of radius \(R_{1}\) is \({B_1} = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{{{R_1}}}\frac{\pi }{2} = \frac{{{\mu _0}I}}{{8{R_1}}} \oplus \) The magnetic field at \(O\) due to arc of radius\({R_2}\) is \(B_{2}=\dfrac{\mu_{0}}{4 \pi} \dfrac{I}{R_{1}} \dfrac{\pi}{2}=\dfrac{\mu_{0} I}{8 R_{2}} \otimes\) The net magnetic field at \(O\) is \({B_1} + {B_2}\) \(B = \frac{{{\mu _0}I}}{{8{R_1}}} + \frac{{{\mu _0}I}}{{8{R_2}}} = \frac{{{\mu _0}I}}{8}\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right) = \frac{{{\mu _0}I}}{8}\left( {\frac{{{R_2} + {R_1}}}{{{R_1}{R_2}}}} \right)\)
KCET - 2012
PHXII04:MOVING CHARGES AND MAGNETISM
362590
A long wire having a semicircular loop of radius \(r\) carries a current \(i\) as shown in figure. The magnetic induction at the centre \(O\) due to entire wire is
1 \(\dfrac{\mu_{0} i}{4 r}\)
2 \(\frac{{{\mu _0}{i^2}}}{{4{r^2}}}\)
3 \(\frac{{{\mu _0}i}}{{4r}}\)
4 None of these
Explanation:
According to Biot-Savart's law, the magnetic induction at a point to a current carrying element \(i \delta l\) is given by \(\delta B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{i\delta 1 \times r}}{{{r^3}}}\) or \(B=\dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l \sin \theta}{r^{2}}\) Directed normal to plane containing \(\delta 1\) and \(r,\theta \) being angle between \(\delta 1\) and \(r\). Field due to semicircular arc Now, angle between a current element \(\delta 1\) of semicircular arc and the radius vector of the element to point \(c\) is \(\pi / 2\). Therefore, the magnitude of magnetic induction \(B\) at \(O\) due to this element is \(\delta B=\dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l \sin \pi / 2}{r^{2}}=\dfrac{\mu_{0} i \delta l}{4 \pi r^{2}}\) Hence, magnetic induction due to whole semicircular loop is \(\begin{aligned}B & =\Sigma \delta B=\Sigma \dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l}{r^{2}} \\& =\dfrac{\mu_{0}}{4 \pi} \dfrac{i}{r^{2}} \Sigma \delta l \\& =\dfrac{\mu_{0} i}{4 \pi r^{2}}(\pi r)=\dfrac{\mu_{0} i}{4 r}\end{aligned}\) The magnetic field due to \(a b\) and \(d e\) is zero, because \(\theta=0^{\circ}\) or \(180^{\circ}\). So, net magnetic field is \(B=\dfrac{\mu_{0} i}{4 r}\)
362587
A circular wire of radius \(R\) is made with uniform material. A current enters at \(A\) and leaves from \(B\). The magnetic field at the centre \(C\) is
1 \(\dfrac{\mu_{0} I}{2 R}\)
2 \(\dfrac{\mu_{0} I}{4 R}\)
3 \(\dfrac{\mu_{0} I}{R}\)
4 Zero
Explanation:
\(\overrightarrow {{B_1}} + \overrightarrow {{B_2}} = 0\) \({B_1} \to \) due to left part \({B_2} \to \) due to right part \(B_{1}, B_{2}\) at centre are numerically equal \(\&\) opposite to each other. (i,e. one is into the paper and the other is towards the reader)
PHXII04:MOVING CHARGES AND MAGNETISM
362588
A wire bent as shown in the figure carries a current \(I\). The magnetic field at \(P\) is
1 \(\dfrac{3 \mu_{0} I}{2 R}\)
2 \(\dfrac{\mu_{0} I}{4 R}\)
3 \(\dfrac{7 \mu_{0} I}{8 R}\)
4 \(\dfrac{\mu_{0} I}{8 R}\)
Explanation:
Let us divide the given conductor into 3 parts. \(B\) due to straight parts \(\left(B_{1}, B_{2}\right)\) is zero and \(B\) due arc is \(\begin{gathered}B_{2}=\left(\dfrac{\mu_{o} I}{2 R} \dfrac{\pi / 2}{2 \pi}\right) \\B=B_{1}+B_{2}+B_{3}=0+\dfrac{\mu_{0} I}{2 R}\left(\dfrac{\pi / 2}{2 \pi}\right)+0=\dfrac{\mu_{0} I}{8 R}\end{gathered}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362589
In the loop shown, the magnetic induction at the point \('O'\) is
Since the point \(O\) lies on the lines of the straight wires, they contribute no magnetic induction at \(O\). The magnetic induction at \(\mathrm{O}\) due to arcs only. The magnetic induction at \(O\) due to are of radius \(R_{1}\) is \({B_1} = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{{{R_1}}}\frac{\pi }{2} = \frac{{{\mu _0}I}}{{8{R_1}}} \oplus \) The magnetic field at \(O\) due to arc of radius\({R_2}\) is \(B_{2}=\dfrac{\mu_{0}}{4 \pi} \dfrac{I}{R_{1}} \dfrac{\pi}{2}=\dfrac{\mu_{0} I}{8 R_{2}} \otimes\) The net magnetic field at \(O\) is \({B_1} + {B_2}\) \(B = \frac{{{\mu _0}I}}{{8{R_1}}} + \frac{{{\mu _0}I}}{{8{R_2}}} = \frac{{{\mu _0}I}}{8}\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right) = \frac{{{\mu _0}I}}{8}\left( {\frac{{{R_2} + {R_1}}}{{{R_1}{R_2}}}} \right)\)
KCET - 2012
PHXII04:MOVING CHARGES AND MAGNETISM
362590
A long wire having a semicircular loop of radius \(r\) carries a current \(i\) as shown in figure. The magnetic induction at the centre \(O\) due to entire wire is
1 \(\dfrac{\mu_{0} i}{4 r}\)
2 \(\frac{{{\mu _0}{i^2}}}{{4{r^2}}}\)
3 \(\frac{{{\mu _0}i}}{{4r}}\)
4 None of these
Explanation:
According to Biot-Savart's law, the magnetic induction at a point to a current carrying element \(i \delta l\) is given by \(\delta B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{i\delta 1 \times r}}{{{r^3}}}\) or \(B=\dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l \sin \theta}{r^{2}}\) Directed normal to plane containing \(\delta 1\) and \(r,\theta \) being angle between \(\delta 1\) and \(r\). Field due to semicircular arc Now, angle between a current element \(\delta 1\) of semicircular arc and the radius vector of the element to point \(c\) is \(\pi / 2\). Therefore, the magnitude of magnetic induction \(B\) at \(O\) due to this element is \(\delta B=\dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l \sin \pi / 2}{r^{2}}=\dfrac{\mu_{0} i \delta l}{4 \pi r^{2}}\) Hence, magnetic induction due to whole semicircular loop is \(\begin{aligned}B & =\Sigma \delta B=\Sigma \dfrac{\mu_{0}}{4 \pi} \cdot \dfrac{i \delta l}{r^{2}} \\& =\dfrac{\mu_{0}}{4 \pi} \dfrac{i}{r^{2}} \Sigma \delta l \\& =\dfrac{\mu_{0} i}{4 \pi r^{2}}(\pi r)=\dfrac{\mu_{0} i}{4 r}\end{aligned}\) The magnetic field due to \(a b\) and \(d e\) is zero, because \(\theta=0^{\circ}\) or \(180^{\circ}\). So, net magnetic field is \(B=\dfrac{\mu_{0} i}{4 r}\)
