Absolutely! Let's go through the steps and explanations for each part of the problem shown in your image. --- ### (a) **Magnetic Moment and Torque** #### **Step 1: Magnetic Dipole Moment of a Square Coil** The **magnetic dipole moment** is given by: \[ \vec{\mu} = N I A \hat{n} \] where: - \( N \) = number of turns (here, \( N = 1 \)) - \( I \) = current - \( A \) = area of the coil (\( a^2 \) for a square of side \( a \)) - \( \hat{n} \) = unit normal vector to the plane For this problem: - \( N = 1 \) - \( A = a^2 \) - The normal's direction (using the right-hand rule) is given as negative \( y \)-direction: \( \hat{n} = -\hat{j} \) So, \[ \vec{\mu} = I a^2 (-\hat{j}) = -I a^2 \hat{j} \] #### **Step 2: Torque on the Coil** The **torque** on a current loop in a magnetic field is: \[ \vec{\tau} = \vec{\mu} \times \vec{B} \] If \( \vec{B} \) is along the \( x \)-direction and \( \vec{\mu} \) along the \( -y \)-direction: \[ \vec{\tau} = -I a^2 \hat{j} \times B \hat{i} = 0 \] (Because \( \hat{j} \times \hat{i} = -\hat{k} \), but if \( \vec{B} \) has no component in the plane of the loop, torque is zero.) **Conclusion:** - **Magnetic moment:** \( \vec{\mu} = -I a^2 \hat{j} \) - **Torque:** \( \vec{\tau} = 0 \) (for this field orientation) --- ### (b) **Magnetic Flux through the Square Coil** #### **Step 1: Magnetic Flux Definition** \[ \Phi_B = \int_S \vec{B} \cdot d\vec{A} \] - The field \( B(x) \) is not constant and depends on \( x \). - The area element \( d\vec{A} \) points out of the plane (normal). #### **Step 2: Integrating the Flux** - Choose the surface to be flat in the \( xz \)-plane. - \( B(x) \) is along \( y \), \( d\vec{A} \) is along \( y \). Set up the flux through the coil of side \( a \): \[ \Phi_B = \int_{0}^{a} dx \int_{0}^{a} dz \, B(x) \] But since \( B \) depends only on \( x \): \[ \Phi_B = a \int_{l_1}^{l_1+a} B(x) dx \] Assuming \( B(x) = \frac{\mu_0 I}{2\pi x} \): \[ \Phi_B = a \int_{l_1}^{l_1+a} \frac{\mu_0 I}{2\pi x} dx = a \frac{\mu_0 I}{2\pi} \int_{l_1}^{l_1+a} \frac{1}{x} dx = a \frac{\mu_0 I}{2\pi} \left[\ln(x)\right]_{l_1}^{l_1+a} \] \[ = a \frac{\mu_0 I}{2\pi} \ln\left(\frac{l_1+a}{l_1}\right) \] If \( l_1 = a \), this becomes: \[ \Phi_B = \frac{\mu_0 I a}{2\pi} \ln(2) \] --- ### (c) **Induced Current When Current Increases** If \( I \) increases linearly: \( I = \alpha t \) so \( \frac{dI}{dt} = \alpha \). #### **Step 1: Faraday's Law** \[ \mathcal{E} = -\frac{d\Phi_B}{dt} \] \[ \Phi_B = \frac{\mu_0 I a}{2\pi} \ln(2) \] So: \[ \frac{d\Phi_B}{dt} = \frac{\mu_0 a}{2\pi} \ln(2) \frac{dI}{dt} \] \[ \mathcal{E} = -\frac{\mu_0 a}{2\pi} \ln(2) \alpha \] #### **Step 2: Induced current via Ohm's Law** \[ I_{\text{induced}} = \frac{\mathcal{E}}{R} = -\frac{\mu_0 a}{2\pi R} \ln(2) \alpha \] --- ## **Summary Table** | Part | Formula & Final Answer | |------|---------|--------------| | **(a)** | \( \vec{\mu} = -I a^2 \hat{j} \) | \( \tau = 0 \) | | **(b)** | \( \Phi_B = \frac{\mu_0 a I}{2\pi} \ln(2) \) | | | **(c)** | \( I_{\text{induced}} = -\frac{\mu_0 a}{2\pi R} \ln(2) \alpha \) | | Let me know if you want a detailed step-by-step derivation for any specific part!