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aev [14]
3 years ago
5

A 45.5-turn circular coil of radius 4.85 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

.490 T. If the coil carries a current of 26.7 mA, find the magnitude of the maximum possible torque exerted on the coil.

Physics
2 answers:
il63 [147K]3 years ago
3 0

Answer:

4.399 Nm

Explanation:

The maximum Torque on a coil is given as,

τ = BNIA...................... Equation 1

Where τ = Maximum torque exerted on the coil, B = Magnetic Field, N = Number of turns, I = Current, A = Area.

Given: N = 45.5 Turns, B = 0.49 T, I = 26.7 mA = 0.0267 A,

A = πr², Where r = radius of the coil, r= 4.85 cm = 0.0485 m

A = 3.14(0.0485)²

A = 7.39×10⁻³ m².

Substitute into equation 1

τ  = 45.5×0.49×26.7×7.39×10⁻³

τ = 4.399 Nm

igomit [66]3 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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pshichka [43]

Answer:

0.050 m

Explanation:

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B=\frac{\mu_0 I}{2\pi r}

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I is the current in the wire

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And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

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A student investigated how the mass of water in an electric kettle affected the time taken for the water to reach boiling point.
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What is the angular displacement of the second hand on a clock after 59 seconds?
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If an electronin an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T
Anni [7]

Answer:

Explanation:

Given that,

Force is downward I.e negative y-axis

F = -2 × 10^-14 •j N

Magnetic field is westward, +x direction

B = 8.3 × 10^-2 •i T

Charge of an electron

q = 1.6 × 10^-19C

Velocity and it direction?

Force in a magnetic field is given as

F = q(V×B)

Angle between V and B is 270, check attachment

The cross product of velocity and magnetic field

F =qVB•Sin270

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v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)

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From cross product

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i×j = k, j×i = -k

j×k =i, k×j = -i

k×i = j, i×k = -j OR -k×i = -j

Comparing -k×i = -j to given problem

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So, the direction of V is negative z- direction

V = -1.51 × 10^6 •k m/s

6 0
3 years ago
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