A current in a wire increases from 2 a to 6 a. how will the magnetic field 0.01 m from the wire change? it increases to four tim
2 answers:
Hi there!
Recall the equation for the magnetic field of a wire (assuming infinite):
B = Magnetic field (T)
μ₀ = Permeability of Free Space (Tm/A)
i = enclosed current (A)
r = distance from wire (m)
From the equation, we can see that there is a DIRECT relationship between the current and the magnetic field.
Therefore, if the current is increased by a factor of 3, <u>the FIELD will also be increased by a factor of three.</u>
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Incomplete part of the question
A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2 . What is the weight Wb of the ball? Express your answer numerically in newtons.
What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale. Calculate your answer from the quantities given in the problem and express it numerically in newtons.
What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative). Express your answer numerically in newtons.
The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and submerged ball is placed on the scale.
What weight W3 does the scale now show?
Answer:
(a) 2.94 N
(b) 1 N
(c) 2.42 N
(d) 3.42 N
Explanation:
(a)
From the definition of density, it's mass per unit volume hence mass is a product of density and volume. To get weight, we multiply mass by acceleration due to gravity
The weight of the ball is
Where is the density, V is volume and g is acceleration due to gravity
Substituting density for 5000 Kg/m3 and g for 9.8 m/s2 and v for 0.00006 m3 then
(b)
Because the ball is being held up mostly by the rod, the fluid pressure on the bottom of the cylinder is just the same as before.
The scale does not "know" the ball is there at all.
That's why it still reads 1 N.
Therefore, the reading is 1 N
(c)
The buoyant force of the fluid on the ball is equal to the weight of the displaced fluid, namely,
so the force needed for the rod to hold up the ball is 2.94 N - 0.52 N = 2.42 N.
(d)
Now the scale "feels" the weight of the ball,
so the scale reads the weight of the ball
PLUS the weight of the original fluid
MINUS the weight of the fluid that was displaced
= 2.94 N + 1.00 N - 0.52 N = 3.42 N