## A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. the wire fully lies in a

Question

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. the wire fully lies in a magnetic field given by (0.3y)i + (0.4y)j Tesla. what is The magnetic force on the wire
A: (0.15i+0.2j) N
B: (0.0015i+0.002j) N
C: (-6.25E-5 k) N
D:(6.25E-5 k) N
E: None of the above

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5 months 2021-08-15T15:46:03+00:00 1 Answers 2 views 0

## Answers ( )

E: None of the above

Explanation:

Since the force on the wire F = ∫idL × B where i = current in wire = 2.0 mA = 2.0 × 10⁻³ A, dL = vector length of wire = dyj and we integrate from y₁ = 0 m and y₂ = 0.25 m, and B = magnetic field = (0.3y)i + (0.4y)j

So, F = ∫idL × B

and we integrate from y₁ = 0 m and y₂ = 0.25 m

F = 2.0 × 10⁻³ A∫₀⁰°²⁵[dyj m × (0.3y)i + (0.4y)j]

F = 2.0 × 10⁻³ A∫₀⁰°²⁵[dyj m × (0.3y)i + dyj m × (0.4y)j]

F = 2.0 × 10⁻³ A∫₀⁰°²⁵[(0.3y) × dy(j × i) + 0.25 × (0.4y)(j × j)]

F = 2.0 × 10⁻³ A∫₀⁰°²°⁵[(0.3ydy(-k) + 0.1y)(0)]

F = 2.0 × 10⁻³ A∫₀⁰°²⁵(-0.3ydy)k

F = 2.0 × 10⁻³ A[-0.3y²/2]₀⁰°²⁵k

F = -2.0 × 10⁻³ A[0.3(0.25)²/2 – 0.3(0)²/2]k

F = -2.0 × 10⁻³ A[0.3(0.0625)/2 – 0]k

F = -2.0 × 10⁻³ A[0.3(0.0625)/2]k

F = -1 × 10⁻³[0.01875]k N

F = -[0.01875 × 10⁻³]k N

F = -[1.875 × 10⁻⁵]k N

Since the answer is not contained in the options, the answer is E.