91Ó°ÊÓ

The term magnetic field may be defined as the area around the magnet behave like a magnet.

The magnetic field can be calculated by using this formula

$dB=\frac{RI}{AN}dt$

Integrate on both sides from$t=0-2.00s$

$$\begin{gathered} \mathrm{Δ}{B}_{0→2}=\frac{R}{AN}\underset{0}{\overset{2}{∫}}Idt \\ =\frac{R}{AN}Area \\ =\frac{R}{AN}\frac{1}{2}\left(2.00s\right)\left(3.00mA\right) \end{gathered}$$

Put value

$$\begin{gathered} \mathrm{Δ}{B}_{0→2}=\frac{\left(0.00300A.s\right)R}{AN} \\ =\frac{\left(0.250\mathrm{Î\copyright }\right)\left(0.00300A.s\right)}{\mathrm{Ï€}{\left(0.00800m\right)}^2\left(4\right)} \\ =0.9325T \end{gathered}$$

Hence, the magnetic field at the location of the for t =2.00s is$\mathrm{Δ}{B}_{0→2}=0.9325T$

Now for t =5.00s

The magnetic field can be calculated as

$$\begin{gathered} \mathrm{Δ}{B}_{0→5}=\mathrm{Δ}{B}_{0→2}+\mathrm{Δ}{B}_{2→5} \\ \mathrm{Δ}{B}_{2→5}=\frac{R\left(Area\right)}{AN} \\ \mathrm{Δ}{B}_{2→5}=\frac{\left(0.250\mathrm{Î\copyright }\right)\left(3.00×{10}^{-3}A\right)\left(3.0s\right)}{\mathrm{Ï€}\left(0.008\right)} \\ \mathrm{Δ}{B}_{2→5}=2.798T \end{gathered}$$

Using value from above result

$$\begin{gathered} \mathrm{Δ}{B}_{0→5}=0.9325T+2.798T \\ \mathrm{Δ}{B}_{0→5}=3.73T \end{gathered}$$

Hence, the magnetic field at the location of the for t = 6.00s islocalid="1658495078469" $\mathrm{Δ}{B}_{0→6}=4.20T$

Now for t =6.00s

The magnetic field can be calculated as

$$\begin{gathered} \mathrm{Δ}{B}_{0→6}=\mathrm{Δ}{B}_{0→5}+\mathrm{Δ}{B}_{5→6} \\ \mathrm{Δ}{B}_{5→6}=\frac{1}{2}\mathrm{Δ}{B}_{0→2} \end{gathered}$$

From graph conclude that area under the curve from t = 0 - 2 is twice of the area under the curve from t = 5 - 6

The magnetic field can be calculated as

$\mathrm{Δ}{B}_{0→6}=\mathrm{Δ}{B}_{0→5}+\frac{1}{2}\mathrm{Δ}{B}_{0→2}$

Put values

$$\begin{gathered} \mathrm{Δ}{B}_{0→6}=3.73T+\frac{1}{2}\left(0.9325T\right) \\ \mathrm{Δ}{B}_{0→6}=4.20T \end{gathered}$$

Hence,the magnetic field at the location of the for is .