Question

A straight wire carries a current of 3 A. Calculate the
magnitude of the magnetic field, at a point, 0.15 m away from the wire.Draw the diagram to show the magnetic field. ​

Answer 1

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Given : Staright wire carries current of 3A

to find : Magnitude of magnetic field at a point 0.15 m away from the wire.

Theory :

• Magnetic field due to current carrying element ( Biot sarvat law )

${\purple{\boxed{\large{\bold{B =</p><p>\frac{u_{0} }{4\pi} \times \frac{i \: dl \: \sin( \alpha ) }{r {}^{2} } }}}}}$

⇒ Special case :

• for a infinte long current carrying wire.

$B= \frac{u_{0}i }{2\pi \: r}$

⇒Value of permeability of free space

$\frac{ u_{0} }{4\pi} = 10 {}^{ - 7} \: henry \: meter \: {}^{ - 1}$

Solution :

I = 3A

r = 0.15 m

$B = \frac{u _{0}i}{2\pi \: r}$

put the values ;

$B= \frac{ u_{0} \times 3}{2\pi \: \times 0.15}$

$B = \frac{u _{0} \times 3 \times 2}{2 \times 2 \pi \: \times 0.15}$

$B= \frac{10 {}^{ - 7} \times 3 \times 2 }{0.15}$

$B= \frac{10 {}^{ - 7} \times 3 \times 2 \times 100}{15}$

$B= 40 \times 10 {}^{ - 7}$

$B = 4 \times 10 {}^{ - 6} \: wbm {}^{ - 2}$

it is the required solution!

⇒For finding the direction of magnetic field

• Right hand thumb rule
• Right hand grip rule .