Physics Moving Charges and Magnetism

A positively charged particle of specific charge α, accelerated by a potential difference V moves through a uniform transverse magnetic field B. The field occupies a region of width l. The angle θ through which the charged particle deviates from initial direction of its motion is

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Vishal Baghel

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p²/2m = Vq => p = √2mVq
θ = sin? ¹ (l/R') = l/R = mV/qB = p/qB

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Kindly consider the following

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Kindly go through the solution

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

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$| \overset{?}{F} | = | I (\overset{?}{L} * \overset{?}{B}) |$

$\begin{array}{r} = | I [L \hat{i} * (2 \hat{i} + 3 \hat{j} - 4 \hat{k})] | \\ = 5 I L \end{array}$

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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:

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${\overset{?}{B}}_{P} = {\overset{?}{B}}_{upper wire} \otimes + {\overset{?}{B}}_{semi-circle} ? + {\overset{?}{B}}_{lower-wire} \otimes$

$B_{P} = - \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 R} - \frac{μ_{0} i}{4 π R} = \frac{μ_{0} i}{4 R} (1 - \frac{2}{π})$ pointing away from the page

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The net magnetic flux through any closed surface is:

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$?_{s} \overset{?}{B} \cdot d \overset{?}{A} = 0$ (No monopole exist)

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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:

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Kindly go through the solution

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Match List-I with List-II

Choose the correct answer from the options given below :

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Gravitational constant $= [M^{- 1} L^{3} T^{- 2}]$

Gravitational potential energy $= [{M L}^{2} T^{- 2}]$

Gravitational potential $= [L^{2} T^{- 2}]$

Gravitational intensity $= [{L T}^{- 2}]$

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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 $20 c m$ at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

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Contributor-Level 10

$f = e v (\frac{μ_{0} i}{2 π r})$

$f = \frac{1.6 * 10^{- 19} * 10^{5} * 2 * 10^{- 7} * 5}{0.2}$

$f = 8 * 10^{- 20}$ Newton

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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:

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New answer posted

a month ago

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:

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