Question

If a = 1.0 cm, b = 3.0 cm, and i = 30 a, what is the magnitude of the magnetic field at point p?

Answer 1

Answer:

Hence the magnitude for magnetic fied at point p at 3 cm will be 15.7*10^-3 T

Step-by-step explanation:

Given;

Path of current for a conductor with two ends distance as 1 cm and 3 cm

Current=30 A.

To Find:

magnitude of the magnetic field at point p

Solution:

magnitude of the magnetic field at point p=B is given by ,

$B=UI/4pie*r^2$ ]*$\int\limits^a_b {r} \, dx$

Here U  and pie are constants, here a=90 degrees and b=0 degree.

$U=4pie*10^-7$  and r=3cm

Therefore ,

$B=[30*10^-7/3*10^-4]*(pie/2)$

$B=15.7*10^-3$

Hence the magnitude for magnetic fied at point p at 3 cm will be 15.7*10^-3

Answer 2

The net magnetic force is $31.4*10^-^5T$

Data;

• a = 1.0cm
• b = 3.0cm
• i = 30a
• B = ?

Magnitude of Magnetic Field at point P

The magnetic field at point 2 will be

$B_2 = \frac{\mu_o I}{4\pi R_2} * \theta \\B_2 = \frac{\mu_o I}{4\pi R_2} * \frac{\pi}{2} = 47.1 * 10^-^5T$

The magnetic field due at point 4 will be

$B_4 = \frac{\mu_o I}{4\pi R_2} * \theta \\B_4 = \frac{\mu_o I}{4\pi R_2} * \frac{\pi}{4}\\B_4 = 15.7*10^-^5T$

The direction of magnetic field due to 2 and 4 are in the opposite direction.

The Net magnetic force is

$B = 47.1*10^-^5 - 15.7*10^-^5 = 31.4*10^-^5T$

The net magnetic force is $31.4*10^-^5T$

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