Question

A non-uniform magnetic field points out of the page as shown below in the figure. The field increases at a constant rate of 2.0 mT as you move to the right. A square wire loop of 15 cm on a side lies on a plane perpendicular to the field, and a 2.5 A current circles the loop in the counter-clockwise direction. Calculate the magnitude and direction of the net magnetic force on the loop

out 15 cm .

Answer 1

The torque on a current loop is given by

The torque on a current loop is given by T = IBA sin emptyset Here I = 2.5 A B = 2.0 mT = 2.0 times 10^- 3 T A = 0.15 times 0.15 = 0.0225 m^2 Here it is given that the coil is perpendicular to the magnetic field so emptyset = 90^0 So we have T = 2.5 times 2.0 times 10^- 3 times 0.0225 times sin 90^0 = 2.5 times 2.0 times 10^- 3 times 0.0225 times 1 = 0.0001125 N - m Direction is perpendicular to both the magnetic field and area of the coil.

Here

$B=2.0 \ mT = 2.0 \times 10^{-3} \ T$

$A=0.15 \times 0.15=0.0225 \ m^2$

Here it is given that the coil is perpendicular to the magnetic field so $\phi = 90^0$

So we have

$\tau = 2.5 \times 2.0 \times 10^{-3} \times 0.0225 \times \sin 90^0 = 2.5 \times 2.0 \times 10^{-3} \times 0.0225 \times 1=0.0001125 \ N-m$

Direction is perpendicular to both the magnetic field and area of the coil.