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

Q 1 :
A long straight wire of length 2 m and mass 250 g is suspended horizontally in a uniform horizontal magnetic field of 0.7 T. The amount of current flowing through the wire will be ( $g = 9.8$ $m$ $s^{- 2}$ ) [2023]

2.45 A
2.25 A
2.75 A
1.75 A

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(4)

[IMAGE 205]

Weight of the wire, $F = m g$ ...(i)

Force acting on the current carrying wire, $F = I (\vec{l} \times \vec{B})$ ...(ii)

Equating eq(i) and (ii)

$m g = I l B \sin 90 °$

$m g = I l B$

$I = \frac{m g}{l B} = \frac{250 \times 10^{- 3} \times 9.8}{2 \times 0.7}$

$I = 1.75$ $A$

Q 2 :
A wire carrying a current $I$ along the positive $x$ -axis has length $L$ . It is kept in a magnetic field $\vec{B} = (2 \hat{i} + 3 \hat{j} - 4 \hat{k})$ $T$ . The magnitude of the magnetic force acting on the wire is [2023]

$5 I L$
$\sqrt{3} I L$
$3 I L$
$\sqrt{5} I L$

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(1)

Given $\vec{B} = (2 \hat{i} + 3 \hat{j} - 4 \hat{k})$ $T$

$\vec{F} = I \vec{L} \times \vec{B} = I L \hat{i} \times (2 \hat{i} + 3 \hat{j} - 4 \hat{k})$ $;$ $\vec{F} = 3 I L \hat{k} + 4 I L \hat{j}; F = 5 I L$

Q 3 :
In the product $\vec{F} = q (\vec{v} \times \vec{B}) = q \vec{v} \times (B \hat{i} + B \hat{j} + B_{0} \hat{k})$

For $q = 1$ and $\vec{v} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$ and $\vec{F} = 4 \hat{i} - 20 \hat{j} + 12 \hat{k}$

What will be the complete expression for $\vec{B}$ ? [2021]

$6 \hat{i} + 6 \hat{j} - 8 \hat{k}$
$- 8 \hat{i} - 8 \hat{j} - 6 \hat{k}$
$- 6 \hat{i} - 6 \hat{j} - 8 \hat{k}$
$8 \hat{i} + 8 \hat{j} - 6 \hat{k}$

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(3)

Given: $\vec{v} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$ and $q = 1; \vec{B} = B \hat{i} + B \hat{j} + B_{0} \hat{k}$

$\vec{F} = 4 \hat{i} - 20 \hat{j} + 12 \hat{k}$

Now, $\vec{v} \times \vec{B} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 6 \\ B & B & B_{0} \end{matrix} |$

$= \hat{i} (4 B_{0} - 6 B) - \hat{j} (2 B_{0} - 6 B) + \hat{k} (2 B - 4 B)$

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

$4 \hat{i} - 20 \hat{j} + 12 \hat{k} = 1$ $[(4 B_{0} - 6 B) \hat{i} - (2 B_{0} - 6 B) \hat{j} - 2 B \hat{k}]$

By comparison:

$4 B_{0} - 6 B = 4$ ...(i)

$- 2 B = 12$

$\Rightarrow B = - 6$ ...(ii)

From eqn. (i) and (ii), $4 B_{0} - 6 (- 6) = 4$

$\Rightarrow 4 B_{0} + 36 = 4 \Rightarrow 4 B_{0} = - 32 \Rightarrow B_{0} = - 8$

So, $\vec{B} = - 6 \hat{i} - 6 \hat{j} - 8 \hat{k}$

Q 4 :
A metallic rod of mass per unit length $0.5$ $kg$ $m^{- 1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30 °$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25$ $T$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is [2018]

7.14 A
5.98 A
14.76 A
11.32 A

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(4)

[IMAGE 206]

Mass per unit length of a metallic rod is $\frac{m}{l} = 0.5$ $kg$ $m^{- 1} .$

Let $I$ be the current flowing.

For equilibrium, $m g \sin 30 ° = I l B \cos 30 °$

$I = \frac{m g}{l B} \tan 30 ° = \frac{0.5 \times 9.8}{0.25 \times \sqrt{3}} = 11.32$ $A$

Topic Question Set

Q 1 :
A long straight wire of length 2 m and mass 250 g is suspended horizontally in a uniform horizontal magnetic field of 0.7 T. The amount of current flowing through the wire will be ( $g = 9.8$ $m$ $s^{- 2}$ ) [2023]

Q 2 :
A wire carrying a current $I$ along the positive $x$ -axis has length $L$ . It is kept in a magnetic field $\vec{B} = (2 \hat{i} + 3 \hat{j} - 4 \hat{k})$ $T$ . The magnitude of the magnetic force acting on the wire is [2023]

Q 3 :
In the product $\vec{F} = q (\vec{v} \times \vec{B}) = q \vec{v} \times (B \hat{i} + B \hat{j} + B_{0} \hat{k})$

For $q = 1$ and $\vec{v} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$ and $\vec{F} = 4 \hat{i} - 20 \hat{j} + 12 \hat{k}$

What will be the complete expression for $\vec{B}$ ? [2021]

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Topic Question Set

Q 1 : A long straight wire of length 2 m and mass 250 g is suspended horizontally in a uniform horizontal magnetic field of 0.7 T. The amount of current flowing through the wire will be (g=9.8 m s-2) [2023]

Q 2 : A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field B→=(2i^+3j^-4k^)T. The magnitude of the magnetic force acting on the wire is [2023]

Q 3 : In the product F→=q(v→×B→)=qv→×(Bi^+Bj^+B0k^) For q=1 and v→=2i^+4j^+6k^ and F→=4i^-20j^+12k^ What will be the complete expression for B→ ? [2021]

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Q 1 :
A long straight wire of length 2 m and mass 250 g is suspended horizontally in a uniform horizontal magnetic field of 0.7 T. The amount of current flowing through the wire will be ( $g = 9.8$ $m$ $s^{- 2}$ ) [2023]

Q 2 :
A wire carrying a current $I$ along the positive $x$ -axis has length $L$ . It is kept in a magnetic field $\vec{B} = (2 \hat{i} + 3 \hat{j} - 4 \hat{k})$ $T$ . The magnitude of the magnetic force acting on the wire is [2023]

Q 3 :
In the product $\vec{F} = q (\vec{v} \times \vec{B}) = q \vec{v} \times (B \hat{i} + B \hat{j} + B_{0} \hat{k})$

For $q = 1$ and $\vec{v} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$ and $\vec{F} = 4 \hat{i} - 20 \hat{j} + 12 \hat{k}$

What will be the complete expression for $\vec{B}$ ? [2021]