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Question: The magnetic field \(d \vec { B }\) due to a small current element \(d \vec { l }\) at a distance ...

The magnetic field dBd \vec { B } due to a small current element dld \vec { l } at a distance r\vec { r } and element carrying current i is,

Or

Vector form of Biot-savart's law is

A

dB=μ04πi(dl×rr)d \vec { B } = \frac { \mu _ { 0 } } { 4 \pi } i \left( \frac { d \vec { l } \times \vec { r } } { r } \right)

B

dB=μ04πi2(dl×rr)d \vec { B } = \frac { \mu _ { 0 } } { 4 \pi } i ^ { 2 } \left( \frac { d \vec { l } \times \vec { r } } { r } \right)

C

dB=μ04πi2(dl×rr2)d \vec { B } = \frac { \mu _ { 0 } } { 4 \pi } i ^ { 2 } \left( \frac { d \vec { l } \times \vec { r } } { r ^ { 2 } } \right)

D

dB=μ04πi(dl×rr3)d \vec { B } = \frac { \mu _ { 0 } } { 4 \pi } i \left( \frac { d \vec { l } \times \vec { r } } { r ^ { 3 } } \right)

Answer

dB=μ04πi(dl×rr3)d \vec { B } = \frac { \mu _ { 0 } } { 4 \pi } i \left( \frac { d \vec { l } \times \vec { r } } { r ^ { 3 } } \right)

Explanation

Solution

dB=μ04πidlsinθr2d B = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i d l \sin \theta } { r ^ { 2 } }dB=μ04πi(dl×r)r3d \vec { B } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i ( d \vec { l } \times \vec { r } ) } { r ^ { 3 } }