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A long wire lying along the x-axis carries a current of 1.20 A in the +x direction. There is a uniform magnetic field present, given by B=0.003i + 0.004j + 0.002k, where i, j, k are the unit vectors along the cartesian coordinate axes') in units of Tesla. Calculate the y-component of the magnetic force acting on a segment of wire of length L = 12.0 cm.

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Answer:

F = (-0.000288ĵ + 0.000576k) N

The y-component for the force = -0.000288 N

Step-by-step explanation:

The magnetic force on a current carrying wire is given as the vector product of current vector and the magnetic field vector.

F = IL × B

I = 1.2î

L = 12.0 cm = 0.12 m

IL = 1.2 × 0.12 = 0.144î

B = (0.003i + 0.004j + 0.002k) T

F = (0.144î) × (0.003i + 0.004j + 0.002k)

F =

| î k j |

|0.144 0 0 |

|0.003 0.004 0.002|

F = î(0 - 0) - ĵ[(0.002 × 0.144) - 0] + k[(0.144×0.004) - 0]

F = (0î - 0.000288ĵ + 0.000576k) N

F = (-0.000288ĵ + 0.000576k) N

The y-component for the force = -0.000288 N

Hope this Helps!!!

User Selman Ulug
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