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A large metal sheet carries an electric current along its surface with $ \lambda$ being the current per unit length. The value of the magnetic field near the surface of the sheet is

$\begin {array} {1 1} (a)\;\mu_0 \lambda & \quad (b)\;\large\frac{\mu_0 \lambda}{2} \\ (c)\;\large\frac{\mu_0 \lambda}{2\pi} & \quad (d)\;\large\frac{\mu_0 \lambda}{\pi} \end {array}$


1 Answer

Construct an ampere’s loop of small width and length $ℓ$. The current passing through the loop is $ \lambda$
$2B ℓ = \mu_0 (\lambda\: ℓ) \rightarrow B = \large\frac {\mu_0 \lambda}{2}$
Ans : (b)
answered Feb 19, 2014 by thanvigandhi_1
edited Mar 14, 2014 by balaji.thirumalai

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