NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153246
A length of wire carries a steady current. It is bent first to form a circular coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is
1 double of its first value
2 quarter of its first value
3 four times of its first value
4 same as the first value
Explanation:
C Given, $\mathrm{N}_{1}=1, \mathrm{~N}_{2}=2$ Let the radius be $r_{1}$ and $r_{2}$ respectively. $2 \pi \mathrm{r}_{1}=2 \times 2 \pi \mathrm{r}_{2}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=2$ Magnetic field $\mathrm{B}$ at the centre of the coil radius $\mathrm{r}_{1}$, $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{~N}_{1} \mathrm{I}}{2 \mathrm{r}_{1}}$ Magnetic field $\mathrm{B}$ at the center of the coil of radius $\mathrm{r}_{2}$, $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{~N}_{2} \mathrm{I}}{2 \mathrm{r}_{2}}$ On dividing equation (ii) by (i), we get $\text { So, } \frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}} =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \times \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\mathrm{~B}_{2} =\mathrm{B}_{1} \times \frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \times \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\mathrm{~B}_{2} =\mathrm{B}_{1} \times \frac{2}{1} \times 2$ $\mathrm{~B}_{2} =4 \mathrm{~B}_{1}$ Magnetic field is four times of its first value.
Manipal UGET - 2016
Moving Charges & Magnetism
153249
In the figure shown, the magnetic field induction at the point $O$ will be
B According to question, We know field induction due to straight wire of infinite length $=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$ and field induction for semicircle $=\frac{\mu_{0} \pi \mathrm{i}}{4 \pi \mathrm{r}}$ Then, Magnetic field due to different part are $B =B_{a}+B_{b}+B_{c}$ $=\frac{\mu_{0}}{4 \pi} \frac{i}{r}+\frac{\mu_{0} \pi i}{4 \pi r}+\frac{\mu_{0} i}{4 \pi r}$ $=\left(\frac{\mu_{0}}{4 \pi}\right)\left(\frac{i}{r}\right)(\pi+2)$
Manipal UGET-2010
Moving Charges & Magnetism
153251
A circular current carrying coil has a radius $R$. The distance from the centre of the coil on the axis of the coil, where the magnetic induction is $1 / 8$ th of its value at the centre of the coil is
1 $\sqrt{3} \mathrm{R}$
2 $\mathrm{R} / \sqrt{3}$
3 $\left(\frac{2}{\sqrt{3}}\right) \mathrm{R}$
4 $\frac{\mathrm{R}}{2 \sqrt{3}}$
Explanation:
A Let, $\mathrm{Z}$ is the distance from the centre of the coil on the axis of the coil. Magnetic field on the axis of circular coil $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{iR}}{2\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{\frac{3}{2}}}$ Magnetic field at the center of circular coil $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}$ Given, $\mathrm{B}_{1}=\frac{1}{8} \mathrm{~B}_{2}$ $\frac{\mu_{0} i \mathrm{R}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3 / 2}}=\frac{1}{8} \frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}$ $8 \mathrm{R}^{3}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3 / 2}$ $\left(8 \mathrm{R}^{3}\right)^{2}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3}$ $\left(4 \mathrm{R}^{2}\right)^{3}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3}$ $4 \mathrm{R}^{2}=\mathrm{R}^{2}+\mathrm{Z}^{2}$ $\mathrm{Z}^{2}=3 \mathrm{R}^{2}$ $\mathrm{Z}=\sqrt{3} \mathrm{R}$
Manipal UGET-2010
Moving Charges & Magnetism
153252
For the figure, the magnetic field at a point $p$ will be
1 $\frac{\mu_{0}}{4} \pi$
2 $\frac{\mu_{0}}{\pi} \otimes$
3 $\frac{\mu_{0}}{2 \pi} \otimes$
4 $\frac{\mu_{0}}{2} \pi$
Explanation:
C Given, $\mathrm{I}_{1}=5 \mathrm{~A}$ $\mathrm{I}_{2}=2.5 \mathrm{~A}$ The magnetic field due to a infinite wire carrying current I at a point lie at a distance $\mathrm{P}$ from the wire is given by $\mathrm{B}==\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}$ For wire $1, \mathrm{~d}_{1}=2.5 \mathrm{~m}$ Thus magnetic field at point $\mathrm{P}$ due to current $\mathrm{I}_{1}$, $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \pi \mathrm{d}_{1}}$ $\mathrm{~B}_{1}=\frac{\mu_{0}(5)}{2 \pi \times(2.5)}=\frac{\mu_{0}}{\pi} \quad \text { (into the paper) }$ For wire $2, \mathrm{~d}_{2}=2.5 \mathrm{~m}$ The magnetic field at point $\mathrm{P}$ due to $\mathrm{I}_{2}$, $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \pi \mathrm{d}_{2}}$ $\mathrm{~B}_{2}=\frac{\mu_{0} \times 2.5}{2 \pi(2.5)}=\frac{\mu_{0}}{2 \pi} \quad \text { (out of the paper) }$ Net magnetic field $\mathrm{B}=\mathrm{B}_{1}-\mathrm{B}_{2}$ $\mathrm{~B}=\frac{\mu_{0}}{\pi}-\frac{\mu_{0}}{2 \pi}$ $\mathrm{B}=\frac{\mu_{0}}{2 \pi} \otimes$
153246
A length of wire carries a steady current. It is bent first to form a circular coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is
1 double of its first value
2 quarter of its first value
3 four times of its first value
4 same as the first value
Explanation:
C Given, $\mathrm{N}_{1}=1, \mathrm{~N}_{2}=2$ Let the radius be $r_{1}$ and $r_{2}$ respectively. $2 \pi \mathrm{r}_{1}=2 \times 2 \pi \mathrm{r}_{2}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=2$ Magnetic field $\mathrm{B}$ at the centre of the coil radius $\mathrm{r}_{1}$, $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{~N}_{1} \mathrm{I}}{2 \mathrm{r}_{1}}$ Magnetic field $\mathrm{B}$ at the center of the coil of radius $\mathrm{r}_{2}$, $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{~N}_{2} \mathrm{I}}{2 \mathrm{r}_{2}}$ On dividing equation (ii) by (i), we get $\text { So, } \frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}} =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \times \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\mathrm{~B}_{2} =\mathrm{B}_{1} \times \frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \times \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\mathrm{~B}_{2} =\mathrm{B}_{1} \times \frac{2}{1} \times 2$ $\mathrm{~B}_{2} =4 \mathrm{~B}_{1}$ Magnetic field is four times of its first value.
Manipal UGET - 2016
Moving Charges & Magnetism
153249
In the figure shown, the magnetic field induction at the point $O$ will be
B According to question, We know field induction due to straight wire of infinite length $=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$ and field induction for semicircle $=\frac{\mu_{0} \pi \mathrm{i}}{4 \pi \mathrm{r}}$ Then, Magnetic field due to different part are $B =B_{a}+B_{b}+B_{c}$ $=\frac{\mu_{0}}{4 \pi} \frac{i}{r}+\frac{\mu_{0} \pi i}{4 \pi r}+\frac{\mu_{0} i}{4 \pi r}$ $=\left(\frac{\mu_{0}}{4 \pi}\right)\left(\frac{i}{r}\right)(\pi+2)$
Manipal UGET-2010
Moving Charges & Magnetism
153251
A circular current carrying coil has a radius $R$. The distance from the centre of the coil on the axis of the coil, where the magnetic induction is $1 / 8$ th of its value at the centre of the coil is
1 $\sqrt{3} \mathrm{R}$
2 $\mathrm{R} / \sqrt{3}$
3 $\left(\frac{2}{\sqrt{3}}\right) \mathrm{R}$
4 $\frac{\mathrm{R}}{2 \sqrt{3}}$
Explanation:
A Let, $\mathrm{Z}$ is the distance from the centre of the coil on the axis of the coil. Magnetic field on the axis of circular coil $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{iR}}{2\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{\frac{3}{2}}}$ Magnetic field at the center of circular coil $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}$ Given, $\mathrm{B}_{1}=\frac{1}{8} \mathrm{~B}_{2}$ $\frac{\mu_{0} i \mathrm{R}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3 / 2}}=\frac{1}{8} \frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}$ $8 \mathrm{R}^{3}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3 / 2}$ $\left(8 \mathrm{R}^{3}\right)^{2}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3}$ $\left(4 \mathrm{R}^{2}\right)^{3}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3}$ $4 \mathrm{R}^{2}=\mathrm{R}^{2}+\mathrm{Z}^{2}$ $\mathrm{Z}^{2}=3 \mathrm{R}^{2}$ $\mathrm{Z}=\sqrt{3} \mathrm{R}$
Manipal UGET-2010
Moving Charges & Magnetism
153252
For the figure, the magnetic field at a point $p$ will be
1 $\frac{\mu_{0}}{4} \pi$
2 $\frac{\mu_{0}}{\pi} \otimes$
3 $\frac{\mu_{0}}{2 \pi} \otimes$
4 $\frac{\mu_{0}}{2} \pi$
Explanation:
C Given, $\mathrm{I}_{1}=5 \mathrm{~A}$ $\mathrm{I}_{2}=2.5 \mathrm{~A}$ The magnetic field due to a infinite wire carrying current I at a point lie at a distance $\mathrm{P}$ from the wire is given by $\mathrm{B}==\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}$ For wire $1, \mathrm{~d}_{1}=2.5 \mathrm{~m}$ Thus magnetic field at point $\mathrm{P}$ due to current $\mathrm{I}_{1}$, $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \pi \mathrm{d}_{1}}$ $\mathrm{~B}_{1}=\frac{\mu_{0}(5)}{2 \pi \times(2.5)}=\frac{\mu_{0}}{\pi} \quad \text { (into the paper) }$ For wire $2, \mathrm{~d}_{2}=2.5 \mathrm{~m}$ The magnetic field at point $\mathrm{P}$ due to $\mathrm{I}_{2}$, $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \pi \mathrm{d}_{2}}$ $\mathrm{~B}_{2}=\frac{\mu_{0} \times 2.5}{2 \pi(2.5)}=\frac{\mu_{0}}{2 \pi} \quad \text { (out of the paper) }$ Net magnetic field $\mathrm{B}=\mathrm{B}_{1}-\mathrm{B}_{2}$ $\mathrm{~B}=\frac{\mu_{0}}{\pi}-\frac{\mu_{0}}{2 \pi}$ $\mathrm{B}=\frac{\mu_{0}}{2 \pi} \otimes$
153246
A length of wire carries a steady current. It is bent first to form a circular coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is
1 double of its first value
2 quarter of its first value
3 four times of its first value
4 same as the first value
Explanation:
C Given, $\mathrm{N}_{1}=1, \mathrm{~N}_{2}=2$ Let the radius be $r_{1}$ and $r_{2}$ respectively. $2 \pi \mathrm{r}_{1}=2 \times 2 \pi \mathrm{r}_{2}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=2$ Magnetic field $\mathrm{B}$ at the centre of the coil radius $\mathrm{r}_{1}$, $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{~N}_{1} \mathrm{I}}{2 \mathrm{r}_{1}}$ Magnetic field $\mathrm{B}$ at the center of the coil of radius $\mathrm{r}_{2}$, $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{~N}_{2} \mathrm{I}}{2 \mathrm{r}_{2}}$ On dividing equation (ii) by (i), we get $\text { So, } \frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}} =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \times \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\mathrm{~B}_{2} =\mathrm{B}_{1} \times \frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \times \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\mathrm{~B}_{2} =\mathrm{B}_{1} \times \frac{2}{1} \times 2$ $\mathrm{~B}_{2} =4 \mathrm{~B}_{1}$ Magnetic field is four times of its first value.
Manipal UGET - 2016
Moving Charges & Magnetism
153249
In the figure shown, the magnetic field induction at the point $O$ will be
B According to question, We know field induction due to straight wire of infinite length $=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$ and field induction for semicircle $=\frac{\mu_{0} \pi \mathrm{i}}{4 \pi \mathrm{r}}$ Then, Magnetic field due to different part are $B =B_{a}+B_{b}+B_{c}$ $=\frac{\mu_{0}}{4 \pi} \frac{i}{r}+\frac{\mu_{0} \pi i}{4 \pi r}+\frac{\mu_{0} i}{4 \pi r}$ $=\left(\frac{\mu_{0}}{4 \pi}\right)\left(\frac{i}{r}\right)(\pi+2)$
Manipal UGET-2010
Moving Charges & Magnetism
153251
A circular current carrying coil has a radius $R$. The distance from the centre of the coil on the axis of the coil, where the magnetic induction is $1 / 8$ th of its value at the centre of the coil is
1 $\sqrt{3} \mathrm{R}$
2 $\mathrm{R} / \sqrt{3}$
3 $\left(\frac{2}{\sqrt{3}}\right) \mathrm{R}$
4 $\frac{\mathrm{R}}{2 \sqrt{3}}$
Explanation:
A Let, $\mathrm{Z}$ is the distance from the centre of the coil on the axis of the coil. Magnetic field on the axis of circular coil $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{iR}}{2\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{\frac{3}{2}}}$ Magnetic field at the center of circular coil $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}$ Given, $\mathrm{B}_{1}=\frac{1}{8} \mathrm{~B}_{2}$ $\frac{\mu_{0} i \mathrm{R}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3 / 2}}=\frac{1}{8} \frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}$ $8 \mathrm{R}^{3}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3 / 2}$ $\left(8 \mathrm{R}^{3}\right)^{2}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3}$ $\left(4 \mathrm{R}^{2}\right)^{3}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3}$ $4 \mathrm{R}^{2}=\mathrm{R}^{2}+\mathrm{Z}^{2}$ $\mathrm{Z}^{2}=3 \mathrm{R}^{2}$ $\mathrm{Z}=\sqrt{3} \mathrm{R}$
Manipal UGET-2010
Moving Charges & Magnetism
153252
For the figure, the magnetic field at a point $p$ will be
1 $\frac{\mu_{0}}{4} \pi$
2 $\frac{\mu_{0}}{\pi} \otimes$
3 $\frac{\mu_{0}}{2 \pi} \otimes$
4 $\frac{\mu_{0}}{2} \pi$
Explanation:
C Given, $\mathrm{I}_{1}=5 \mathrm{~A}$ $\mathrm{I}_{2}=2.5 \mathrm{~A}$ The magnetic field due to a infinite wire carrying current I at a point lie at a distance $\mathrm{P}$ from the wire is given by $\mathrm{B}==\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}$ For wire $1, \mathrm{~d}_{1}=2.5 \mathrm{~m}$ Thus magnetic field at point $\mathrm{P}$ due to current $\mathrm{I}_{1}$, $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \pi \mathrm{d}_{1}}$ $\mathrm{~B}_{1}=\frac{\mu_{0}(5)}{2 \pi \times(2.5)}=\frac{\mu_{0}}{\pi} \quad \text { (into the paper) }$ For wire $2, \mathrm{~d}_{2}=2.5 \mathrm{~m}$ The magnetic field at point $\mathrm{P}$ due to $\mathrm{I}_{2}$, $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \pi \mathrm{d}_{2}}$ $\mathrm{~B}_{2}=\frac{\mu_{0} \times 2.5}{2 \pi(2.5)}=\frac{\mu_{0}}{2 \pi} \quad \text { (out of the paper) }$ Net magnetic field $\mathrm{B}=\mathrm{B}_{1}-\mathrm{B}_{2}$ $\mathrm{~B}=\frac{\mu_{0}}{\pi}-\frac{\mu_{0}}{2 \pi}$ $\mathrm{B}=\frac{\mu_{0}}{2 \pi} \otimes$
153246
A length of wire carries a steady current. It is bent first to form a circular coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is
1 double of its first value
2 quarter of its first value
3 four times of its first value
4 same as the first value
Explanation:
C Given, $\mathrm{N}_{1}=1, \mathrm{~N}_{2}=2$ Let the radius be $r_{1}$ and $r_{2}$ respectively. $2 \pi \mathrm{r}_{1}=2 \times 2 \pi \mathrm{r}_{2}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=2$ Magnetic field $\mathrm{B}$ at the centre of the coil radius $\mathrm{r}_{1}$, $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{~N}_{1} \mathrm{I}}{2 \mathrm{r}_{1}}$ Magnetic field $\mathrm{B}$ at the center of the coil of radius $\mathrm{r}_{2}$, $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{~N}_{2} \mathrm{I}}{2 \mathrm{r}_{2}}$ On dividing equation (ii) by (i), we get $\text { So, } \frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}} =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \times \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\mathrm{~B}_{2} =\mathrm{B}_{1} \times \frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \times \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\mathrm{~B}_{2} =\mathrm{B}_{1} \times \frac{2}{1} \times 2$ $\mathrm{~B}_{2} =4 \mathrm{~B}_{1}$ Magnetic field is four times of its first value.
Manipal UGET - 2016
Moving Charges & Magnetism
153249
In the figure shown, the magnetic field induction at the point $O$ will be
B According to question, We know field induction due to straight wire of infinite length $=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$ and field induction for semicircle $=\frac{\mu_{0} \pi \mathrm{i}}{4 \pi \mathrm{r}}$ Then, Magnetic field due to different part are $B =B_{a}+B_{b}+B_{c}$ $=\frac{\mu_{0}}{4 \pi} \frac{i}{r}+\frac{\mu_{0} \pi i}{4 \pi r}+\frac{\mu_{0} i}{4 \pi r}$ $=\left(\frac{\mu_{0}}{4 \pi}\right)\left(\frac{i}{r}\right)(\pi+2)$
Manipal UGET-2010
Moving Charges & Magnetism
153251
A circular current carrying coil has a radius $R$. The distance from the centre of the coil on the axis of the coil, where the magnetic induction is $1 / 8$ th of its value at the centre of the coil is
1 $\sqrt{3} \mathrm{R}$
2 $\mathrm{R} / \sqrt{3}$
3 $\left(\frac{2}{\sqrt{3}}\right) \mathrm{R}$
4 $\frac{\mathrm{R}}{2 \sqrt{3}}$
Explanation:
A Let, $\mathrm{Z}$ is the distance from the centre of the coil on the axis of the coil. Magnetic field on the axis of circular coil $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{iR}}{2\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{\frac{3}{2}}}$ Magnetic field at the center of circular coil $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}$ Given, $\mathrm{B}_{1}=\frac{1}{8} \mathrm{~B}_{2}$ $\frac{\mu_{0} i \mathrm{R}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3 / 2}}=\frac{1}{8} \frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}$ $8 \mathrm{R}^{3}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3 / 2}$ $\left(8 \mathrm{R}^{3}\right)^{2}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3}$ $\left(4 \mathrm{R}^{2}\right)^{3}=\left(\mathrm{R}^{2}+\mathrm{Z}^{2}\right)^{3}$ $4 \mathrm{R}^{2}=\mathrm{R}^{2}+\mathrm{Z}^{2}$ $\mathrm{Z}^{2}=3 \mathrm{R}^{2}$ $\mathrm{Z}=\sqrt{3} \mathrm{R}$
Manipal UGET-2010
Moving Charges & Magnetism
153252
For the figure, the magnetic field at a point $p$ will be
1 $\frac{\mu_{0}}{4} \pi$
2 $\frac{\mu_{0}}{\pi} \otimes$
3 $\frac{\mu_{0}}{2 \pi} \otimes$
4 $\frac{\mu_{0}}{2} \pi$
Explanation:
C Given, $\mathrm{I}_{1}=5 \mathrm{~A}$ $\mathrm{I}_{2}=2.5 \mathrm{~A}$ The magnetic field due to a infinite wire carrying current I at a point lie at a distance $\mathrm{P}$ from the wire is given by $\mathrm{B}==\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}$ For wire $1, \mathrm{~d}_{1}=2.5 \mathrm{~m}$ Thus magnetic field at point $\mathrm{P}$ due to current $\mathrm{I}_{1}$, $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \pi \mathrm{d}_{1}}$ $\mathrm{~B}_{1}=\frac{\mu_{0}(5)}{2 \pi \times(2.5)}=\frac{\mu_{0}}{\pi} \quad \text { (into the paper) }$ For wire $2, \mathrm{~d}_{2}=2.5 \mathrm{~m}$ The magnetic field at point $\mathrm{P}$ due to $\mathrm{I}_{2}$, $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \pi \mathrm{d}_{2}}$ $\mathrm{~B}_{2}=\frac{\mu_{0} \times 2.5}{2 \pi(2.5)}=\frac{\mu_{0}}{2 \pi} \quad \text { (out of the paper) }$ Net magnetic field $\mathrm{B}=\mathrm{B}_{1}-\mathrm{B}_{2}$ $\mathrm{~B}=\frac{\mu_{0}}{\pi}-\frac{\mu_{0}}{2 \pi}$ $\mathrm{B}=\frac{\mu_{0}}{2 \pi} \otimes$
