A bar magnet moves through a loop of wire with constant velocity, and the north pole enters the loop first the induced current w
2 answers:
Answer:
Loop must be near the end of the pole of bar magnet
Explanation:
As we know by Faraday's law of electromagnetic induction the induced EMF or induced current in the loop is given by
here we know that
now we know that as the magnet comes closer to the loop the magnetic field due to any of the pole of magnet will increase.
So here the maximum flux will pass through the magnet when magnet is closer to the loop.
So here when A bar magnet moves through a loop of wire with constant velocity, and the north pole enters the loop first the induced current will be greatest when the magnet is located so that the loop is when Loop must be near the end of the pole of bar magnet
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Answer:
The charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C
Explanation:
Here is the complete question
Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.
Solution
The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by
F = kq₁q₂/r₁²
q₁q₂ = Fr₁²/k
q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²
q₁q₂ = 6 × 10⁻¹⁶ C² (1)
When the charges are brought together, the charge is now q = (q₁ + q₂)/2
The new repulsive force F = 1.406 × 10⁻⁴ N at a distance of r₂ = 20 cm = 0.2 m is then
F₂ = kq²/r₂²
q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²
q² = 6.25 × 10⁻¹⁶ C²
q = √(6.25 × 10⁻¹⁶) C
q = 2.5 × 10⁻⁸ C
(q₁ + q₂)/2 = 2.5 × 10⁻⁸ C
(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C
q₁ + q₂ = 5 × 10⁻⁸ C (2)
q₁ = 5 × 10⁻⁸ C - q₂ (3)
Substituting equation (3) into (1), we have
(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²
Expanding the bracket, we have
(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²
So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0
Using the quadratic formula to find q₂
q₁ = 5 × 10⁻⁸ C - q₂
q₁ = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C
q₁ = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C
So the charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C