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3 years ago

How would a small bar magnet be oriented when placed at position x?

Physics

2 answers:

algol [13]3 years ago

8 0

Answer:

Explanation:

As per the convention the Magnetic field lines starts from the North pole. As can be seen the diagram the magnetic field is closed and as per the direction of field lines, right side of the magnet will be North pole and left side will be south pole. When one will keep a magnet at X, it will align itself with the magnetic field of the magnet. Since opposite poles attract, the magnet will align as per the orientation shown in option D.

jonny [76]3 years ago

5 0

The answer is D.

The shapes of the magnetic field lines of a bar magnet are closed. By convention, the direction of the field is taken out of the north pole and entering through the south pole of the magnet.

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A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$

For part (c)

The electric field according to Gauss Law is given by:

$EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C$

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Ann [662]

Answer:

Explanation:

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When an object is fully magnetized, all of its magnetic domains will be ?

kkurt [141]

The answer is to this question D

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Gala2k [10]

Answer:

e. all of these

Explanation:

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Alenkasestr [34]

Answer:

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