A current-carrying wire 0.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10.0 A and there
is a resultant force of 3.0 N on the wire due to the interaction of the current and field, what is the magnetic field strength?
1 answer:
<em>Answer</em>
0.6 teslas
<em>Explanation</em>
When a conductor is inside a magnetic field it experiences a force given by;
Force = ILBsinθ
Where I⇒ current
L ⇒length of the conductor
B ⇒ magnetic field strength
θ ⇒ Angle between the conductor and magnetic field.
F = ILBsinθ
When θ = 90°, Then sin 90 =1 and the formula becomes;
F =ILB
3 = 10 × 0.5 × B
3 = 5B
B = 3/5
= 0.6
magnetic field strength = 0.6 teslas
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