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A 2.2 m -long wire carries a current of 7.2 a and is immersed within a uniform magnetic field b⃗. when this wire lies along the x axis, a magnetic force f⃗=(−2.6j)n acts on the wire, and when it lies on the y axis, the force is f⃗=(2.6i−5.0k) the x-component of b⃗ .

User EoLithic
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Final Answer:

The magnetic force acting on the 2.2 m long wire is 10.9 N.

Explanation:

To determine the magnetic force acting on the wire, we need to use the formula:

F = B × I

where F is the force, B is the magnetic field strength, and I is the current flowing through the wire.

First, let’s find the magnetic field strength. The magnetic field is uniform, so we can use the equation:

B = μ₀I / 2l

where μ₀ is the permeability of free space (4π x 10^-7 T·m/A), I is the current flowing through the wire (7.2 A), and l is the length of the wire (2.2 m).

Plugging in the values, we get:

B = μ₀I / 2l = (4π x 10^-7 T·m/A) x (7.2 A) / (2.2 m) = 0.089 T

Next, we can find the x-component of the magnetic force by multiplying the magnetic field strength by the current:

Fx = B × I = 0.089 T x 7.2 A = 6.3 N

The y-component of the magnetic force can be found by using the equation:

Fy = B × I = 0.089 T x 7.2 A = 4.6 N

Finally, we add the x- and y-components to find the total magnetic force:

F = Fx + Fy = 6.3 N + 4.6 N = 10.9 N

Therefore, the magnetic force acting on the 2.2 m long wire is 10.9 N.

User Clint Warner
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