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- ¯¯¯¯¯¯¯δB=μ04πdl(→l×→r)r3
- ¯¯¯¯¯¯¯δB=μ04πI(→dl×→r)r3
- ¯¯¯¯¯¯¯δB=μ04πI(→r×→dl)r3
- ¯¯¯¯¯¯¯δB=μ04πI(→dl×→r)r2

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Solution

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→δB=μ04πIdlsin(θ)r2^r, where θ is the angle between the line element →dl and the radial unit vector ^r.

Now we know: ^r=→rr, using this and the cross product of two vectors we get; →δB=μ04πI(→dl×→r)r3

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