NEET PYQ: Moving Charges and Magnetism Important Questions with Solutions

Q 1 :
From Ampere’s circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is [2022]

uniform and remains constant for both the regions
a linearly increasing function of distance up to the boundary of the wire and then linearly decreasing for the outside region
a linearly increasing function of distance $r$ up to the boundary of the wire and then decreasing one with $1 / r$ dependence for the outside region
a linearly decreasing function of distance up to the boundary of the wire and then a linearly increasing one for the outside region

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(3)

Using Ampere’s circuital law

$B_{in} = \frac{μ_{0} I r}{2 π a^{2}}, or B_{in} \propto r$

$B_{out} = \frac{μ_{0} I}{2 π r}, or B_{out} \propto \frac{1}{r}$

Q 2 :
A thick current carrying cable of radius $‘ R ’$ carries current $‘ I ’$ uniformly distributed across its cross-section. The variation of magnetic field $B (r)$ due to the cable with the distance $‘ r ’$ from the axis of the cable is represented by [2021]

[IMAGE 212]
[IMAGE 213]
[IMAGE 214]
[IMAGE 215]

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(4)

[IMAGE 216]

By Ampere's circuital law,

$\oint \vec{B} \cdot d \vec{l} = μ_{0} \times current losed by path$

$\Rightarrow B \cdot 2 π r = μ_{0} \times \frac{I r^{2}}{a^{2}} (∵ R = a)$

$\Rightarrow B = \frac{μ_{0} I r}{2 π a^{2}} (for r < a)$

At surface $r = a,$ so $B = \frac{μ_{0} I}{2 π a}$

$∴$ $B = \frac{μ_{0} I}{2 π r} (for r > a)$

The variation of magnetic field with distance $' r'$ from the axis is given by the above figure.

Q 3 :
An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of $10^{5}$ $m/s$ parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

Electron $v = 10^{5}$ $m/s$ [2021]

[IMAGE 217]

$8 \times 10^{- 20}$ $N$
$4 \times 10^{- 20}$ $N$
$8 π \times 10^{- 20}$ $N$
$4 π \times 10^{- 20}$ $N$

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(1)

[IMAGE 218]

The magnetic field due to a straight wire at the location of the electron is

$B = \frac{μ_{0}}{4 π} \cdot \frac{2 I}{r}$

$B = 10^{- 7} \times \frac{2 \times 5}{0.2} = 50 \times 10^{- 7} T (\hat{k})$

The direction is given by right-hand thumb rule.

The force on a charged particle moving in a magnetic field is

$\vec{F} = q (\vec{v} \times \vec{B})$

$\vec{F} = - 1.6 \times 10^{- 19} \times (10^{5} \hat{i} \times 50 \times 10^{- 7} \hat{k})$

$\vec{F} = 8 \times 10^{- 20} N \hat{j}$

Q 4 :
A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field, B with the distance, d from the centre of the conductor, is correctly represented by the figure [2019]

[IMAGE 219]
[IMAGE 220]
[IMAGE 221]
[IMAGE 222]

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(4)

Magnetic field due to long solid cylindrical conductor of radius $R$ ,

(i) For $d < R,$ $I' = \frac{I d^{2}}{R^{2}}$

$\int \vec{B} \cdot d \vec{l} = μ_{0} I' \Rightarrow B (2 π d) = \frac{μ_{0} I d^{2}}{R^{2}} \Rightarrow B = \frac{μ_{0} I d}{2 π R^{2}}$

$∴$ $B \propto d$

(ii) For $d = R,$ $B = \frac{μ_{0} I}{2 π R}$ (maximum)

(iii) For $d > R,$ $B = \frac{μ_{0} I}{2 π d} \Rightarrow B \propto \frac{1}{d}$

Q 5 :
A long straight wire of radius $a$ carries a steady current $I$ . The current is uniformly distributed over its cross-section. The ratio of the magnetic fields $B$ and $B'$ , at radial distances $a / 2$ and $2 a$ respectively, from the axis of the wire is [2016]

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$\frac{1}{4}$
$\frac{1}{2}$

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(1)

Magnetic field at a point inside the wire at distance $r = (\frac{a}{2})$ from the axis of wire is

$B = \frac{μ_{0} I}{2 π a^{2}} r = \frac{μ_{0} I}{2 π a^{2}} \times \frac{a}{2} = \frac{μ_{0} I}{4 π a}$

Magnetic field at a point outside the wire at a distance $r = 2 a$ from the axis of wire is

$B' = \frac{μ_{0} I}{2 π r} = \frac{μ_{0} I}{2 π} \times \frac{1}{2 a} = \frac{μ_{0} I}{4 π a}$ $∴$ $\frac{B}{B'} = 1$

Q 6 :
Two identical long conducting wires $A O B$ and $C O D$ are placed at right angle to each other, with one above other such that $O$ is their common point for the two. The wires carry $I_{1}$ and $I_{2}$ currents, respectively. Point $P$ is lying at distance $d$ from $O$ along a direction perpendicular to the plane containing the wires. The magnetic field at the point $P$ will be [2014]

$\frac{μ_{0}}{2 π d} (\frac{I_{1}}{I_{2}})$
$\frac{μ_{0}}{2 π d} (I_{1} + I_{2})$
$\frac{μ_{0}}{2 π d} (I_{1}^{2} - I_{2}^{2})$
$\frac{μ_{0}}{2 π d} {(I_{1}^{2} + I_{2}^{2})}^{1 / 2}$

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(4)

[IMAGE 223]

The magnetic field at the point $P$ , at a perpendicular distance $d$ from $O$ in a direction perpendicular to the plane $A B C D$ due to currents through $A O B$ and $C O D$ are perpendicular to each other. Hence,

$B = {(B_{1}^{2} + B_{2}^{2})}^{1 / 2}$

$= [{(\frac{μ_{0}}{4 π} \frac{2 I_{1}}{d})}^{2} + {(\frac{μ_{0}}{4 π} \frac{2 I_{2}}{d})}^{2}]^{1 / 2}$

$= \frac{μ_{0}}{2 π d} {(I_{1}^{2} + I_{2}^{2})}^{1 / 2}$

Topic Question Set

Q 1 :
From Ampere’s circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is [2022]

Q 2 :
A thick current carrying cable of radius $‘ R ’$ carries current $‘ I ’$ uniformly distributed across its cross-section. The variation of magnetic field $B (r)$ due to the cable with the distance $‘ r ’$ from the axis of the cable is represented by [2021]

Q 4 :
A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field, B with the distance, d from the centre of the conductor, is correctly represented by the figure [2019]

Q 5 :
A long straight wire of radius $a$ carries a steady current $I$ . The current is uniformly distributed over its cross-section. The ratio of the magnetic fields $B$ and $B'$ , at radial distances $a / 2$ and $2 a$ respectively, from the axis of the wire is [2016]

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Topic Question Set

Q 1 : From Ampere’s circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is [2022]

Q 2 : A thick current carrying cable of radius ‘R’ carries current ‘I’ uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance ‘r’ from the axis of the cable is represented by [2021]

Q 4 : A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field, B with the distance, d from the centre of the conductor, is correctly represented by the figure [2019]

Q 5 : A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B′, at radial distances a/2 and 2a respectively, from the axis of the wire is [2016]

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Q 1 :
From Ampere’s circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is [2022]

Q 2 :
A thick current carrying cable of radius $‘ R ’$ carries current $‘ I ’$ uniformly distributed across its cross-section. The variation of magnetic field $B (r)$ due to the cable with the distance $‘ r ’$ from the axis of the cable is represented by [2021]

Q 4 :
A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field, B with the distance, d from the centre of the conductor, is correctly represented by the figure [2019]

Q 5 :
A long straight wire of radius $a$ carries a steady current $I$ . The current is uniformly distributed over its cross-section. The ratio of the magnetic fields $B$ and $B'$ , at radial distances $a / 2$ and $2 a$ respectively, from the axis of the wire is [2016]