Question

A uniform magnetic field $\overrightarrow{B} = 0.25\hat{k}T$ exists in a circular region of radius R = 5 m. A loop of radius R = 5m lying in x – y plane encloses the magnetic field at t = 0 and then pulled at uniform velocity $\overrightarrow{v} = 4\hat{i}m/s$. Find the emf induced (in volts) is the loop at t = 2 sec.

Answer 1

6

Answer 2

The magnetic flux through the loop is given by $\Phi_B = B \times A_{eff}$, where $A_{eff}$ is the area of intersection between the magnetic field region and the loop. Both are circles of radius $R$. The distance between their centers at time $t$ is $d = vt$. The area of intersection of two circles of radius $R$ with their centers separated by a distance $d$ is $A(d) = 2R^2 \cos^{-1}\left(\frac{d}{2R}\right) - \frac{d}{2}\sqrt{4R^2-d^2}$.

Faraday's law states that the induced EMF is $\mathcal{E} = -\frac{d\Phi_B}{dt}$. Since $\Phi_B(t) = B \times A(vt)$, we have $\mathcal{E} = -B \frac{dA(vt)}{dt}$. The derivative of the area with respect to time is $\frac{dA}{dt} = \frac{dA}{d} \frac{dd}{dt}$, where $\frac{dd}{dt} = v$. The derivative of $A(d)$ with respect to $d$ is $\frac{dA}{d} = \frac{d^2 - 4R^2}{\sqrt{4R^2-d^2}}$. Therefore, $\frac{dA}{dt} = v \left( \frac{(vt)^2 - 4R^2}{\sqrt{4R^2-(vt)^2}} \right) = -v \frac{4R^2-(vt)^2}{\sqrt{4R^2-(vt)^2}} = -v\sqrt{4R^2-(vt)^2}$ for $0 \le vt \le 2R$.

The induced EMF is $\mathcal{E} = -B \left( -v\sqrt{4R^2-(vt)^2} \right) = Bv\sqrt{4R^2-(vt)^2}$.

Given values: $B = 0.25$ T $R = 5$ m $v = 4$ m/s $t = 2$ sec

At $t=2$ sec, $d = vt = 4 \times 2 = 8$ m. Since $d = 8 \le 2R = 10$, the formula is applicable.

Substituting the values: $\mathcal{E} = (0.25 \text{ T}) \times (4 \text{ m/s}) \times \sqrt{4 \times (5 \text{ m})^2 - (8 \text{ m})^2}$ $\mathcal{E} = 1 \text{ V} \times \sqrt{4 \times 25 \text{ m}^2 - 64 \text{ m}^2}$ $\mathcal{E} = \sqrt{100 \text{ m}^2 - 64 \text{ m}^2}$ $\mathcal{E} = \sqrt{36 \text{ m}^2}$ $\mathcal{E} = 6$ V