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3 years ago

9

A 8.0-cm long solenoid has 2000 turns of wire and carries a current of 5.0-A. Calculate the strength of the magnetic field at th

e center of the solenoid in teslas.

1 answer:

CaHeK987 [17]3 years ago

6 0

Answer:

B = 0.157 T

Explanation:

Given that,

Length of the solenoid, l = 8 cm = 0.08 m

Number of turns in the wire, N = 2000

Current, I = 5 A

We need to find the strength of the magnetic field at the center of the solenoid. It is given by the formula as follows :

$N=\mu_o nI$ , N is number of turns per unit length of solenoid.

So,

$B=4\pi \times 10^{-7}\times \dfrac{2000}{0.08}\times 5\\\\=0.157\ T$

So, the magnetic field at the center of the solenoid is 0.157 T.

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A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

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$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

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And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

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$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

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$Q = 0.3 + 0.18$

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$Q_f = KE_f + U_f$

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$KE_f = \frac{1 }{2} m (v_2^2$

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$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

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$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

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2 years ago