A loop of area 0.250 m² is in a uniform 0.020 0-T magnetic field.If the flux through the loop is 3.83 ´ 10^-3 T·m²,what angle does the normal to the plane of the loop make with the direction of the magnetic field? A)40.0 $\degree$ B)50.0 $\degree$ C)37.5 $\degree$ D)This is not possible.

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Quizplus · Accepted Answer

The magnetic flux through a loop is given by:$$\Phi_B=BA\cos\theta$$where:* $\Phi_B$ is the magnetic flux (in webers)* $B$ is the magnetic field strength (in teslas)* $A$ is the area of the loop (in square meters)* $\theta$ is the angle between the normal to the plane of the loop and the direction of the magnetic field (in radians)In this problem, we are given the following information:* $\Phi_B=3.83\times10^{-3}\text{ T}\cdot\text{m}^2$* $B=0.0200\text{ T}$* $A=0.250\text{ m}^2$We can use this information to solve for $\theta$:$$\cos\theta=\frac{\Phi_B}{BA}=\frac{3.83\times10^{-3}\text{ T}\cdot\text{m}^2}{(0.0200\text{ T})(0.250\text{ m}^2)}=0.766$$Taking the inverse cosine of both sides, we get:$$\theta=\cos^{-1}(0.766)=\boxed{40.0^\circ}$$Therefore, the correct answer is A.

A Loop of Area 0 °\degree° B)500 °\degree° C)375 °\degree° D)This Is Not Possible

A Loop of Area 0 $\degree$ B)500 $\degree$ C)375 $\degree$ D)This Is Not Possible