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Pavlova-9 [17]
3 years ago
12

When the north pole of a bar magnet is moved into a solenoid, the needle on a galvanometer attached to the solenoid moves to the

left, indicating that a current is produced. The magnet is then pulled out at a faster speed. What will be the result
Physics
2 answers:
NNADVOKAT [17]3 years ago
7 0

Answer:

The neddle of the galvanometer will move to the right at a faster speed.

Explanation:

Given:

The north pole of a bar magnet is moving towards a solenoid and the needle of the galvanometer moves to the left.

When a bar magnet has a relative deisplacement with respect to a solenoid, then it will induce some emf within the solenoid according to Faraday's law. When the north pole is moving towards the solenoid it will induce a current.

When the bar magnet is moving away from the solenoid at a faster speed, the rate of change of flux in the solenoid is faster. So the induced current will change in  faster way, but in the opposite direction.

Thus, the neddle of the galvanometer will move to the right at a faster speed.

zzz [600]3 years ago
4 0

Answer:

a larger current in the opposite direction

Explanation:

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Answer:

The answer is "4659.2 \times 10^{-24} \ N"

Explanation:

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\to B=\frac{\mu_0 i R^2}{(R^2+x^2)^{\frac{3}{2}}}\\\\

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\to B=\frac{(4\pi\times 10^{-7})(2.80\ A) (\frac{0.35}{2})^2}{( (\frac{0.35}{2})^2+ (\frac{0.24}{2})^2)^{\frac{3}{2}}}\\\\

       =\frac{(12.56 \times 10^{-7})(2.80\ A) \times 0.030625}{( 0.030625+ 0.0144)^{\frac{3}{2}}}\\\\=\frac{  1.07702 \times 10^{-7} }{0.0095538976}\\\\=112.730955 \times 10^{-7}\\\\=1.12\times 10^{-5}\ \ T\\

Calculating the force experienced through the protons:

F=qvB=(1.6 \times 10^{-19}) (2600)(1.12 \times 10^{-5})= 4659.2 \times 10^{-24}\ N

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