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1 month ago

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A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh

en it is at an angle of 30° with respect to the field?

2 answers:

Sav [3.1K]1 month ago

7 0

Answer:

The force of magnetism, F = 0.24 N

Explanation:

The provided information includes:

Current in the wire, I = 4 A

Wire length, L = 20 cm = 0.2 m

Magnetic field strength, B = 0.6 T

Force and magnetic field angle, θ = 30°. The formula for magnetic force is:

$F=ILB\ sin\theta$

$F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)$

F = 0.24 N

Consequently, the force acting on the wire at a 30° angle to the magnetic field is 0.24 N. This completes the solution.

ValentinkaMS [3.4K]1 month ago

7 0

Explanation:

The following data is provided.

Length = 20 cm, current = 4 A

B = 0.6 T, $\theta$ = 30°

Thus, the equation for calculating the force on the wire is as follows.

F = $IlBsin \theta$

Substituting the values into this equation yields:

F = $IlBsin \theta$

= $4 A \times 20 cm \times \frac{10^{-2}}{1 cm} \times 0.6 T \times Sin(30^{o})$

= 0.24 N

= 0.2 N

This leads us to the conclusion that the force acting on the wire is 0.2 N.

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