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Current I si following along the path AB...

Current I si following along the path `ABCD`, along the four edges of the cube (figure-a) creates a magnetic field in the centre at the center of the cube current I flowing along the path of the six edges `ABCDHEA` .

A

`sqrt((3)/(2)) B_(0)` Towards corner `G` .

B

`sqrt3 B_(0)` Towards corner `E`

C

`sqrt((3)/(2)) B_(0)` Towards conrner `H` .

D

`sqrt3 B_(0)` Towards corner `F` .

Text Solution

Verified by Experts

The correct Answer is:
D
Each edge share current for two loops so cur rent for one loop is `(1)/(2)` and its field at centre of loop is `(B_(0))/(2)` There are sin loops
`oversetrarr(B) =((B_(0))/(2) + (B_(0))/(2))hatk`
`oversetrarrB = B_(0) hati + B_(0) hatj +B_(0hat)hatk`
`B =sqrt3 B_(0)` towards `F` .
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