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3 years ago

A circular wire loop of radius 15.0 cm carries a current of 2.60

1 answer:

5 0

Part (a): Magnetic dipole moment

Magnetic dipole moment = IA, I = Current, A = Area of the loop
Then,
Magnetic dipole moment = 2.6*π*0.15^2 = 0.184 Am^2

Part (b): Torque acting on the loop
T = IAB SinФ, where B = Magnetic field, Ф = Angle
Then,
T = Magnetic dipole moment*B*SinФ = 0.184*12*Sin 41 = 1.447 Nm

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2 years ago

If the net force acting on an object is 0 N, you can be sure thr forces acting on the object are,

Bas_tet [7]

Answer:

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Explanation:

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2 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

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3 years ago

Atoms that are bonded together to form a new material with new<br> properties and characteristics. .

mihalych1998 [28]

Answer:

Explanation:

Atoms form chemical bonds to make their outer electron shells more stable. ... An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.

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2 years ago

The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it

soldi70 [24.7K]

Answer:

6.21 m/s

Explanation:

Using work energy equation then

$U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})$

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, $v_a$ as 0, 9.81 for g then

$58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s$

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2 years ago