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3 years ago

Please help me with my question due tomorrow morning,

Physics

1 answer:

lara [203]3 years ago

8 0

Answer:

Explanation:

we just have to add the resistance and the ans comes 14...i am not sure but we do it like this

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3 years ago

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A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$

Hence, the magnetic field is $B=2.958\times10^{-5}T$

7 0

3 years ago

Please help me with my question due tomorrow morning,​

Please help me with my question due tomorrow morning,