Question

A circular loop of radius 0.0400 m is
oriented 75.0 degrees to a magnetic
field. If the magnetic flux through the
loop is 9.59 x 10-7 Wb, what is the
strength of the field?

Answer 1

Let's see

Find enclosed area of loop

$\\ \rm\dashrightarrow \pi r^2$

$\\ \rm\dashrightarrow 3.14(0.04)^2$

$\\ \rm\dashrightarrow 0.005m^2$

Now

$\\ \rm\dashrightarrow \phi=BAcos\theta$

• We need B

$\\ \rm\dashrightarrow 9.59\times 10^{-7}=B(0.005)cos75$

$\\ \rm\dashrightarrow 9.59\times 10^{-7}=0.0046B$

$\\ \rm\dashrightarrow B=2084.78\times 10^{-7}$

$\\ \rm\dashrightarrow B=2.08\times 10^{-4}T$

$\\ \rm\dashrightarrow B=0.208mT$

Answer 2

Answer:7.37*10^-4

Explanation:B=Ф/A*cos(θ) B=9.59*10^7/(π*0.0400²)*cos(75)=7.37144316*10^-4 =7.37*10^-4