(1) Doubling of the current through the wire will result in doubling of its magnetic field.
The magnetic field around a wire is a function of the current I and radial distance r
(with mu denoting the magnetic permeability of the medium). So, B is directly proportional to I. The field magnitude will double with the doubled current from 5A to 10A
(2) Using the same formula as in (1), we can see that the magnetic field is inversely proportional to the radial distance from the wire. So, a particle at 20cm will experience half the magnitude compared to a particle at 10cm.
(3) Answer
If a particle with a charge q moves through a magnetic field B with velocity v, it will be acted on by the magnetic force
So, a particle with charge -2uC will experience a magnetic force of same magnitude but opposite direction (and perpendicular to B) as compared to a particle with a charge of 2uC
Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g
= 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f = 
⇒ T = 4
M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 × 
×
× 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.
Answer:
Two orbitals for their electrons and six in the 2p subshell
Explanation:
Hope this helps :)
Answer:
2.46 eV
Explanation:
It is given that,
The energy of light that fall on the metal = 3.56 eV
The stopping potential of the metal = 1.1 V
We need to find the work function of the metal. It is given by the relation as follows :
W = E-KE ...(1)
Where KE is the kinetic energy of the ejected electron and it is given by :
KE = V×e
= 1.1 eV
Put all the values in formula (1)
W = 3.56 eV - 1.1 eV
= 2.46 eV
Hence, the work function of the metal is 2.46 eV. Hence, the correct option is (c).
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