Question

Two long straight wires A and B are placed 50 cm apart and carry current 20A
and 15A respectively in same direction. A point P is 40 cm from wire A and 30
cm from wire B. What is the magnitude of resultant magnetic field at 'p' in (UT)
unit? (V2 = 1.414)​

Answer 1

The magnitude of resultant magnetic field at P is $1.41\times10^{-5}\ T$.

Explanation:

Given that,

Current in wire A= 20 A

Current in wire B = 15 A

Distance from point A = 40 cm

Distance from point B = 30 cm

We need to calculate the magnetic field for wire A

Using formula of magnetic field

$B = \dfrac{\mu_{0}I}{2\pi r}$

Put the value into the formula

$B_{A}=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times40\times10^{-2}}$

$B_{A}=1.0\times10^{-5}\ T$

We need to calculate the magnetic field for wire B

Put the value into the formula

$B_{A}=\dfrac{4\pi\times10^{-7}\times15}{2\pi\times30\times10^{-2}}$

$B_{A}=1.0\times10^{-5}\ T$

We need to calculate the magnitude of resultant magnetic field

Using horizontal component of magnetic field

Use cosine rule for angle

$B=B_{A}\cos53-B_{B}\cos37$

Put the value in the equation

$B=1.0\times10^{-5}\times\cos53-1.0\times10^{-5}\times\cos37$

$B=−0.00000196\ T$

Using vertical component of magnetic field

$B'=B_{A}\sin53+B_{B}\sin37$

$B'=1.0\times10^{-5}\times\sin53+1.0\times10^{-5}\times\sin37$

$B'=0.000014\ T$

The resultant of magnetic field

$B=\sqrt{(−0.00000196)^2+(0.000014)^2}$

$B=1.41\times10^{-5}\ T$

Hence, The magnitude of resultant magnetic field at P is $1.41\times10^{-5}\ T$.

Learn more :

Topic : magnetic field

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