A 5-turn square loop (10 cm along a side,resistance = 4.0 $\Omega$)is placed in a magnetic field that makes an angle of 30 $\degree$ with the plane of the loop.The magnitude of this field varies with time according to B = 0.50t²,where t is measured in s and B in T.What is the induced current in the coil at t = 4.0 s? A)25 mA B)5.0 mA C)13 mA D)43 mA E)50 mA

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Quizplus · Accepted Answer

The magnetic flux through the loop is given by:$$\Phi = BA\cos	heta$$where:* B is the magnetic field strength* A is the area of the loop* θ is the angle between the magnetic field and the normal to the loopThe induced emf is given by:$$\varepsilon = -\frac{d\Phi}{dt}$$Substituting the given values, we get:$$\varepsilon = -A\frac{d}{dt}(B\cos	heta) = -A\cos	heta\frac{dB}{dt}$$$$\varepsilon = -(0.10 \ m)^2\cos 30\degree\frac{d}{dt}(0.50t^2 \ T/s)$$$$\varepsilon = -(0.0050 \ m^2)(0.866)(1.00 \ T/s)(8.0 \ s)$$$$\varepsilon = -0.0353 \ V$$The induced current is given by:$$I = \frac{\varepsilon}{R} = \frac{-0.0353 \ V}{4.0 \ \Omega} = -8.83 \ mA$$The negative sign indicates that the current flows in the opposite direction to the assumed direction. Therefore, the induced current in the coil at t = 4.0 s is 8.83 mA.

A 5-Turn Square Loop (10 Cm Along a Side,resistance = Ω\OmegaΩ

A 5-Turn Square Loop (10 Cm Along a Side,resistance = $\Omega$