<span>A wire carrying electric current will produce a magnetic field with closed field lines surrounding the wire.Another version of the right hand rules can be used to determine the magnetic field direction from a current—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it. See.<span>The Biot-Savart Law can be used to determine the magnetic field strength from a current segment. For the simple case of an infinite straight current-carrying wire it is reduced to the form <span><span>B=<span><span><span>μ0</span>I</span><span>2πr</span></span></span><span>B=<span><span><span>μ0</span>I</span><span>2πr</span></span></span></span>.</span><span>A more fundamental law than the Biot-Savart law is Ampere ‘s Law, which relates magnetic field and current in a general way. It is written in integral form as <span><span>∮B⋅dl=<span>μ0</span><span>Ienc</span></span><span>∮B⋅dl=<span>μ0</span><span>Ienc</span></span></span>, where Ienc is the enclosed current and μ0 is a constant.</span><span>A current-carrying wire feels a force in the presence of an external magnetic field. It is found to be <span><span>F=Bilsinθ</span><span>F=Bilsinθ</span></span>, where ℓ is the length of the wire, i is the current, and θ is the angle between the current direction and the magnetic field.</span></span>Key Terms<span><span>Biot-Savart Law: An equation that describes the magnetic field generated by an electric current. It relates the magnetic field to the magnitude, direction, length, and proximity of the electric current. The law is valid in the magnetostatic approximation, and is consistent with both Ampère’s circuital law and Gauss’s law for magnetism.</span><span>Ampere’s Law: An equation that relates magnetic fields to electric currents that produce them. Using Ampere’s law, one can determine the magnetic field associated with a given current or current associated with a given magnetic field, providing there is no time changing electric field present.</span></span>
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Answer:
Formula v=u+at
v= 20+(-2×5)
v= 20+(-10)
v= 20-10
v= 10m/s
Hence, the final velocity is 10m/s.
Answer:
5.7 m
Explanation:
AD = length of the ladder = L = 8 m
AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m
AC = distance of person on the ladder from the bottom end = x
W = weight of the ladder = 240 N
= weight of the person = 710 N
F = force by the wall on the ladder
N = normal force by ground on the ladder = ?
Using equilibrium of force along the vertical direction
N = + W
N = 710 + 240
N = 950 N
μ = Coefficient of static friction = 0.55
f =static frictional force on the ladder
Static frictional force is given as
f = μ N
f = (0.55) (950)
f = 522.5 N
Force equation along the horizontal direction is given as
F = f
F = 522.5 N
using equilibrium of torque about point A
F Sin50 (AD) = W Cos50 (AB) + ( Cos50 (AC))
(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)
x = 5.7 m
Diagram A is the correct diagram