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The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
To find the answer, we need to know about the magnetic field inside the solenoid.
<h3>What's the expression of magnetic field inside a solenoid?</h3>
- Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I
- n = no. of turns per unit length and I = current through the solenoid
<h3>What's is the magnetic field inside the solenoid here?</h3>
- Here, n = 290/32cm or 290/0.32 = 906
I= 0.3 A
- So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.
Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
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Answer:
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Answer:
- 1.07 ft
Explanation:
V1 = (-5, 7, 2)
V2 = (3, 1, 2)
Projection of v1 along v2, we use the following formula
=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}
So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4
The magnitude of vector V2 is given by
=
So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft
Thus, the projection of V1 along V2 is - 1.07 ft.
so we need to find the direction of v2
5m
Explanation:
Given parameters:
Weight of object = 50N
Work done in lifting object = 250J
Unknown:
Vertical height = ?
Solution:
The work done on an object is the force applied to lift a body in a specific direction.
Work done = force x distance
Weight is a force in the presence of gravity;
Work done = weight x height of lifting
Height of lifting =
Height of lifting = = 5m
The vertical height through which the object was lifted is 5m
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