Answer:
mu=12Tm^2
Explanation:
the magnetic moment mu of a single loop is given by:

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

hope this helps!!
The portion of the flux leaves the curved surface of the cylinder is 60%.
<h3 /><h3>What are electrons?</h3>
The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.
If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.
If 20% of the flux leave from one end, then another 20% will leave from another end.
So, the net flux through curved surface is
100 -20 -20 = 60%
Thus, the total flux leaves the curved surface of the cylinder is 60%
Learn more about electrons.
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A rotating disc supplied with constant power where the relationship of the angular velocity of the disc and the number of rotations made by the disc is governed by Newton's second law for rotation. This law is specially made for rotating bodies which is extracted from Newton's second law of motion.
Answer: 14 m/s
Explanation:
The speed
of a sound wave is given by the following equation:
Where:
is the wavelength of the sound wave
Hence:
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
2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2
Then,
v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)
v = 1.51 × 10^6 m/s
Direction of the force
Let x be the direction of v
-F•j = v•x × B•i
From cross product
We know that
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
We notice that
-F•j = q ( -V•k × B×i)
So, the direction of V is negative z- direction
V = -1.51 × 10^6 •k m/s