A steady current \(I\) goes through a wire loop PQR having shape of a right angle triangle with \(\mathrm{PQ}=3 x, \mathrm{PR}=4 x\) and \(\mathrm{QR}=5 x\). If the magnitude of the magnetic field at P due to this loop is \(k\left(\frac{\mu_{0} I}{48 \pi x}\right)\), find the value of \(k\).

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Magnetic Effects of Current

A steady current

I

goes through a wire loop PQR having shape of a right angle triangle with

PQ = 3 x, PR = 4 x

and

QR = 5 x

. If the magnitude of the magnetic field at P due to this loop is

k (\frac{μ _{0} I}{48 π x})

, find the value of

k

Answer: 7

The magnetic field at P is due to the two sides PQ and PR (QR is too far to contribute significantly at P). For a straight wire,

B = \frac{μ _{0} I}{4 π d} (cos θ_{1} + cos θ_{2})

, where

d

is the perpendicular distance. The perpendicular distance from P to QR is

\frac{3 \times 4}{5} x = \frac{12 x}{5}

. The field is

B = \frac{μ _{0} I}{4 π \times \frac{12 x}{5}} (cos 53° + cos 37°) = \frac{μ _{0} I}{4 π \times \frac{12 x}{5}} (\frac{3}{5} + \frac{4}{5}) = \frac{μ _{0} I}{4 π \times \frac{12 x}{5}} \times \frac{7}{5} = \frac{7 μ _{0} I}{48 π x}

. Thus

k = 7

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Integer answer