Divide the long wire in small segments. The field due to one segment can be found from Biot- Savart law.
dB=(uo/4pi) Idlxr/r^3. In this expression if we take vector r perpendicular to current element , the cross product in the expression shows that the direction of B is perpendicular to vector r and from symmetry it can be said that field line is circular around the wire segment. This is true for all segment ,because in infinitely long wire all segments are equivalent . Thus, magnetic field lines are concentric around a long straight wire.
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Answer:
Weight (mass) = 16.5 kg
velocity = 0 m/a
acceleration =2.6 m/s^2
displacement = 13.2m
now,
acceleration = velocity/ time
2.6 = 0 / t
t = o / 2.6
t = o
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
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Answer:
Explanation:
Let the charge on proton be q .
energy gain by proton in a field having potential difference of V₀
= V₀ q
Due to gain of energy , its kinetic energy becomes 1/2 m v₀²
where m is mass and v₀ is velocity of proton
V₀ q = 1/2 m v₀²
In the second case , gain of energy in electrical field
= 2 V₀q , if v be the velocity gained in the second case
2 V₀q = 1/2 m v²
1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²
mv² = 2 m v₀²
v = √2 v₀