IIT-JEE 2009 · Paper 2 · Physics Q54

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PhysicsInteger+4 / -1Question 54 of 57

A steady current $I$ goes through a wire loop PQR having shape of a right angle triangle with PQ $=3x$ , PR $=4x$ and QR $=5x$ . If the magnitude of the magnetic field at P due to this loop is $k\left(\dfrac{\mu_0I}{48\pi x}\right)$ , find the value of $k$ .

Your answer

Answer and worked solution

Official answer: 7

Only QR contributes. P is on the lines PQ and PR, so those sides give no field at P. The right angle is at P ( $3^2+4^2=5^2$ ).

Distance to QR. $d=\dfrac{3x\cdot4x}{5x}=\dfrac{12x}5$ . The foot of the perpendicular divides QR into $\frac{9x}5$ and $\frac{16x}5$ , so the angles at P satisfy $\sin\alpha_1=\frac35$ , $\sin\alpha_2=\frac45$ .

Field.
$B=\frac{\mu_0I}{4\pi d}(\sin\alpha_1+\sin\alpha_2)=\frac{5\mu_0I}{48\pi x}\cdot\frac75=7\left(\frac{\mu_0I}{48\pi x}\right).$

Answer: 7

Topics:Biot-Savart law

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IIT-JEE 2009 · Paper 2, Physics question 54