Physics Moving Charges and Magnetism

An electron is moving along +x direction with a velocity of 6 * 10? ms?¹. It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be:

0 Follower 1 View

Contributor-Level 10

F = q (E+v*B). E+v*B=0

0 0

New answer posted

2 months ago

A particle of charge q and mass m is moving with a velocity -vi (v ≠ 0) towards a large screen placed in the Y-Z plane at a distance d. If there is a magnetic field B = B?k, the minimum value of v for which the particle will not hit the screen is:

0 Follower 3 Views

Contributor-Level 10

r = mv/qB?
To not collide, r∴ Vmax = qB? d/m
Note: It should be maximum instead of minimum.

0 0

New answer posted

2 months ago

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 $1.6 * 10^{- 27} k g$ $^{}$ )

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

0 0

New answer posted

2 months ago

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:

0 Follower 5 Views

Contributor-Level 10

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 $v_{y} = v_{0}$

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

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

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

0 0

New answer posted

2 months ago

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

0 Follower 1 View

Contributor-Level 10

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.

0 0

New answer posted

2 months ago

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:

0 Follower 5 Views

Contributor-Level 10

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 graph representation, which is option A based on the formulas.)

0 0

New answer posted

2 months ago

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 :

0 Follower 2 Views

Contributor-Level 10

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

0 0

New answer posted

2 months ago

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.

0 Follower 2 Views

Contributor-Level 10

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

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

0 0

New answer posted

2 months ago

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-------------.

0 Follower 2 Views

Contributor-Level 10

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

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

0 0

New answer posted

2 months ago

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 $4 π * 1 0^{- 7} SI$ units):

0 Follower 4 Views