Physics Moving Charges and Magnetism

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

Proton with kinetic energy of $1 M e V$ moves from south the north. It gets an acceleration of $10^{12} m / s^{2}$ $^{}^{}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{- 27} k g$ $^{}$ )

6 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

A particle of mass $m$ and charge $q$ has an initial velocity $\vec{v} = v_{0} \hat{j}$ . If an electric field $\vec{E} = E_{0} \hat{i}$ and magnetic field $\vec{E} = E_{0} \hat{i}$ act on the particle, its speed will double after a time:

15 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Since magnetic force cannot change the speed. So only electric field which is along $x$ -direction will change the speed along $X$ -direction only.

$v_{x} = \frac{E_{0} q}{m} t$ but $v_{y} = v_{0}$

$2 v_{0} = \sqrt[]{v_{x}^{2} + v_{y}^{2}}$

$4 v_{0}^{2} = \frac{E_{o}^{2} q^{2} t^{2}}{m^{2}} + v_{0}^{2}$

$t = \frac{\sqrt[]{3} m v_{o}}{q E_{o}}$

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

A charge Q is moving dl distance in the magnetic field B. Find the value of work done by B

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

As we know that magnetic force acting on a charge particle will be
F = q (v * B)
W = F.dl
Since force and displacement will be always perpendicular so work done is always zero.

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

Figure A and B show long straight wires of circular cross section (a and b with a < b), carrying current I which is uniformly distributed across the cross section. The magnitude of magnetic field B varies with radius r and can be represented as:

33 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

As b > a
The magnetic field inside the wire (rR) is B = µ? I/ (2πr).
For wire with radius a, B increases linearly to r=a, then decreases. For wire with radius b, B increases linearly to r=b, then decreases. Since a⇒ B? > B?
B? = µ? I / 2πa
B? = µ? I / 2πb
(Note: The question is likely asking for the gr

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

Two ions having same mass have charges in the ratio 1 : 2. They are projected normally in a uniform magnetic field with their speeds in the ratio 2 : 3. The ratio of the radii of their circular trajectories is :

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Given 2q? = q? and v? /v? = 2/3, m? = m?
r = mv/qB
r? /r? = (m? /m? ) * (v? /v? ) * (q? /q? )
r? /r? = (1) * (2/3) * (2) = 4/3

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

A proton, a deuteron and an α particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic force acting on them is ----------- and their speed is ----------- in the ratio.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$v = \frac{p}{m} \Rightarrow v_{p} : v_{d} : v_{α} = 1 : \frac{1}{2} : \frac{1}{4} = 4 : 2 : 1$

$F_{m a g} = q v B \Rightarrow F_{P} : F_{d} : F_{α} = 1 \times 4 : 1 \times 2 : 2 \times 1 = 2 : 1 : 1$

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

Magnetic fields at two points on the axis of a circular coil at a distance of 0.05m 0.2m from the centre are in the ratio 8 : 1. The radius of coil is-------------.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Magnetic field on the axis of a circular coil at distance x from centre, B = $\frac{μ_{0} N i r^{2}}{2 {(r^{2} + x^{2})}^{\frac{3}{2}}}$ $\frac{_{}^{}}{{^{}^{}}^{\frac{}{}}}$

$\frac{{(r^{2} + {(0.2)}^{2})}^{\frac{3}{2}}}{{(r^{2} + {(0.05)}^{2})}^{\frac{3}{2}}} = 8$ $\frac{r^{2} + {(0.2)}^{2}}{r^{2} + {(0.05)}^{2}} = 4 \Rightarrow r^{2} = 0.01 \Rightarrow r = 0.1 m .$