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A wire of length 0.50 m carries a current of 8.0 A. If it experiences a force of 0.40 N, what is the magnetic field strength?

Physics

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Answer

0.10 T0.10 \text{ T}

Step-by-step solution

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here's the solution for question 12:

Given:
Length of wire, L=0.50 mL = 0.50 \text{ m}
Current, I=8.0 AI = 8.0 \text{ A}
Force on the wire, F=0.40 NF = 0.40 \text{ N}
The wire is at right angles to the magnetic field, so θ=90∘\theta = 90^\circ.

  1. Recall the formula for the magnetic force on a current-carrying wire.

    The formula is F=BILsin⁡θF = BIL \sin \theta.

  2. Substitute the given values into the formula.

    Since the wire is at right angles to the magnetic field, sin⁡90∘=1\sin 90^\circ = 1.

    0.40 N=B(8.0 A)(0.50 m)(1)0.40 \text{ N} = B (8.0 \text{ A})(0.50 \text{ m})(1)
  3. Solve for the magnetic field strength, BB.

    0.40 N=B(4.0 A⋅m)0.40 \text{ N} = B (4.0 \text{ A} \cdot \text{m}) B=0.40 N4.0 A⋅mB = \frac{0.40 \text{ N}}{4.0 \text{ A} \cdot \text{m}} B=0.10 TB = 0.10 \text{ T}

    The strength of the magnetic field is 0.10 T0.10 \text{ T}.

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