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Luda [366]
3 years ago
5

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

of the magnetic field is 0.300 TT , what is ττtau, the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field?
Physics
1 answer:
labwork [276]3 years ago
4 0

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

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A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A
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\omega_f = 585.37 \ rev/s

Explanation:

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I \omega_i = I' \omega_f

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3 0
3 years ago
When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5
Scilla [17]

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C   → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C   → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C   → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first  charge , 2nd to 5 th charges can be  written as 2e,  3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of  1.6 x 10⁻¹⁹ C  exists.

3 0
3 years ago
Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis
ch4aika [34]

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to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = \frac{1}{900*10^3 m}

so put here value in equation 1 for 380 nm

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θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 700 × 10^{-9}

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0
3 years ago
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