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Alborosie
3 years ago
8

A current-carrying wire of length 52.0 cm is positioned perpendicular to a uniform magnetic field. If the current is 15.0 A and

it is determined that there is a resultant force of 2.3 N on the wire due to the interaction of the current and field, what is the magnetic field strength?
Physics
1 answer:
nata0808 [166]3 years ago
6 0

Answer:

The magnetic field strength is 0.29 T.

Explanation:

Given that,

Length of current-carrying wire, L = 52 cm = 0.52 m

Current, I = 15 A

Magnetic force, F = 2.3 N

We need to find the magnetic field strength. We know that the magnetic force is given by :

F=ILB

B is magnetic field strength

B=\dfrac{F}{IL}\\\\B=\dfrac{2.3}{0.52\times 15}\\\\B=0.29\ T

So, the magnetic field strength is 0.29 T.

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38.4 m/s

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a) at t = 3.2s. x = 6 * 3.2^2 = 61.44 m

b) at t = 3.2 + Δt. x = 6*(3.2 + \Delta t)^2

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v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}

v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}

v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}

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Answer:

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F = k q₁ q₂ / r²

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Take the square root of both sides:

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