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Magnetism

Section: Physics

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Question : 44 of 46

Marks: +1, -0

A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3, PR = 4x and QR = 5x. 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 Adv 2009 P2]

Your Answer:

Solution:

The segments PQ and PR, in the triangle shown here, cannot produce

\vec{B}

at point P since point P lies on them. Here, only QR creates

\vec{B}

at point P.

Applying

B_P

=

\frac{\mu_0 i}{4\pi R}(\cos\phi_1 + \cos\phi_2)

, where

\phi_1

= 53° and

\phi_2

= 37°, we get the magnitude of the magnetic field as follows:

B_P

=

\frac{\mu_0 I}{4\pi(4\pi \sin 37^{\circ})}

(cos 53° + cos 37°)

=

\frac{\mu_0 I}{16\pi \times \frac{3}{5}} \left( \frac{3x}{5x} + \frac{4x}{5x} \right)

=

\frac{5\mu_0 I}{48\pi x} \left( \frac{7}{5} \right)

=

\frac{7\mu_0 I}{48\pi x}

=

k\left( \frac{\mu_0 I}{48\pi x} \right)

where k = 7.

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