Moving Charges and Magnetism NEET PYQ – Chapter wise Previous Year Questions

Q 1 :

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 π \times 10^{- 7}$ SI units) [2024]

44 mT
4.4 T
4.4 mT
44 T

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(3)

Number of turns in the coil, $N = 100$

Radius of coil, $r = 10$ $cm = 0.1$ $m$

Current flowing in the coil, $I = 7$ $A$

As, $B = \frac{μ_{0} N I}{2 r}$

$B = \frac{4 π \times 10^{- 7} \times 100 \times 7}{2 \times 0.1} = \frac{4 \times 3.14 \times 10^{- 7} \times 100 \times 7}{2 \times 0.1}$

$= 43.96 \times 10^{- 4} = 4.4$ $mT$

Q 2 :

A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by [2023]

$\frac{μ_{0} i}{4 R} [1 - \frac{2}{π}]$ pointed away from page
$\frac{μ_{0} i}{4 R} [1 - \frac{2}{π}]$ pointed into the page
$\frac{μ_{0} i}{4 R}$ pointed into the page
$\frac{μ_{0} i}{4 R}$ pointed away from the page

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(1)

$B_{p}$ due to wire $1 = \frac{μ_{0} i}{4 π R} \otimes$

$B_{p}$ due to wire $3 = \frac{μ_{0} i}{4 π R} \otimes$

$B_{p}$ due to wire $2 = \frac{μ_{0} i}{4 R} ⊙$

$B_{net} = - \frac{μ_{0} i}{2 π R} + \frac{μ_{0} i}{4 R} = \frac{μ_{0} i}{4 R} [- \frac{2}{π} + 1] = \frac{μ_{0} i}{4 R} [1 - \frac{2}{π}]$

Pointed away from page.

Q 3 :

A straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$ . The total magnetic field at the centre $P$ of the loop is [2019]

Zero
$\frac{3 μ_{0} i}{32 R}$ , outward
$\frac{3 μ_{0} i}{32 R}$ , inward
$\frac{μ_{0} i}{2 R}$ , inward

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(1)

Magnetic field due to $i_{1} = \frac{μ_{0} i_{1}}{2 R} \frac{θ_{1}}{2 π}$ (Into the plane)

Magnetic field due to $i_{2} = \frac{μ_{0} i_{2}}{2 R} \frac{θ_{2}}{2 π}$ (Out of the plane)

For parallel combination

Now, $\frac{i_{1}}{i_{2}} = \frac{ρ l_{2}}{A} \times \frac{A}{ρ l_{1}} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{\frac{1}{4} (2 π R)}{\frac{3}{4} (2 π R)} = \frac{1}{3}$

$\Rightarrow i_{1} = \frac{i_{2}}{3} \Rightarrow i_{2} = 3 i_{1}$

$∴$ Net magnetic field, $= \frac{μ_{0} i_{1}}{2 R} (\frac{θ_{1}}{2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{θ_{2}}{2 π})$

$= \frac{μ_{0} i_{1}}{2 R} (\frac{3 π}{2 \times 2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{π}{2 \times 2 π})$

$= \frac{μ_{0}}{2 R} [\frac{3 i_{1}}{4} - \frac{i_{2}}{4}] = \frac{μ_{0}}{2 R} [\frac{3 i_{1}}{4} - \frac{3 i_{1}}{4}] = 0$

Q 4 :

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is $B$ . It is then bent into a circular coil of $n$ turns. The magnetic field at the centre of this coil of $n$ turns will be [2016]

$n B$
$n^{2} B$
$2 n B$
$2 n^{2} B$

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(2)

Let $l$ be the length of the wire. Magnetic field at the centre of the loop is

$B = \frac{μ_{0} I}{2 R} ∴ B = \frac{μ_{0} π I}{l} (∵ l = 2 π R)$ ...(i)

$B' = \frac{μ_{0} n I}{2 r} = \frac{μ_{0} n I}{2 (\frac{l}{2 π n})} or, B' = \frac{μ_{0} n^{2} π I}{l}$ ...(ii)
From eqns. (i) and (ii), we get $B' = n^{2} B$

Q 5 :

A wire carrying current $I$ has the shape as shown in adjoining figure.
Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius $R$ is lying in $Y - Z$ plane. Magnetic field at point O is [2015]

$\vec{B}$ $= - \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} + 2 \hat{k})$
$\vec{B}$ $= \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} - 2 \hat{k})$
$\vec{B}$ $= \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} + 2 \hat{k})$
$\vec{B}$ $= - \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} - 2 \hat{k})$

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(1)

Given situation is shown in the figure.

Parallel wires 1 and 3 are semi-infinite, so magnetic field at O due to them

${\vec{B}}_{1} = {\vec{B}}_{3} = - \frac{μ_{0} I}{4 π R} \hat{k}$

Magnetic field at O due to semi-circular arc in YZ-plane is given by ${\vec{B}}_{2} = - \frac{μ_{0} I}{4 R} \hat{i}$

Net magnetic field at point O is given by

$\vec{B} = {\vec{B}}_{1} + {\vec{B}}_{2} + {\vec{B}}_{3}$

$= - \frac{μ_{0} I}{4 π R} \hat{k} - \frac{μ_{0} I}{4 R} \hat{i} - \frac{μ_{0} I}{4 π R} \hat{k} = - \frac{μ_{0} I}{4 π R} (π \hat{i} + 2 \hat{k})$

Topic Question Set

Q 1 :

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 π \times 10^{- 7}$ SI units) [2024]

Q 2 :

A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by [2023]

Q 3 :

A straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$ . The total magnetic field at the centre $P$ of the loop is [2019]

Q 4 :

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is $B$ . It is then bent into a circular coil of $n$ turns. The magnetic field at the centre of this coil of $n$ turns will be [2016]

Q 5 :

A wire carrying current $I$ has the shape as shown in adjoining figure.
Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius $R$ is lying in $Y - Z$ plane. Magnetic field at point O is [2015]

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Topic Question Set

Q 1 : A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π×10-7 SI units) [2024]

Q 2 : A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by [2023]

Q 3 : A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P of the loop is [2019]

Q 4 : A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be [2016]

Q 5 : A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is [2015]

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Q 1 :

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 π \times 10^{- 7}$ SI units) [2024]

Q 2 :

A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by [2023]

Q 3 :

A straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$ . The total magnetic field at the centre $P$ of the loop is [2019]

Q 4 :

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is $B$ . It is then bent into a circular coil of $n$ turns. The magnetic field at the centre of this coil of $n$ turns will be [2016]

Q 5 :

A wire carrying current $I$ has the shape as shown in adjoining figure.
Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius $R$ is lying in $Y - Z$ plane. Magnetic field at point O is [2015]