1. **Problem statement:** A horseshoe magnet produces a uniform magnetic field of flux density $B$ between its poles. A wire XY of length 4.4 cm is placed horizontally and normal to the magnetic field between the poles. A current of 2.6 A flows from X to Y. The top-pan balance reading increases by 2.3 g. (i) State and explain the polarity of pole P. (ii) Calculate the flux density $B$ between the poles. (c) When the current is replaced by a low frequency sinusoidal current of r.m.s. 2.6 A, calculate the variation in the reading of the top-pan balance. --- 2. **Key formula and rules:** - The magnetic force on a current-carrying wire in a magnetic field is given by: $$F = BIL\sin\theta$$ where $B$ is the flux density, $I$ is the current, $L$ is the length of wire in the field, and $\theta$ is the angle between wire and magnetic field. - Since the wire is normal to the magnetic field, $\sin\theta = 1$. - The force acts perpendicular to both the magnetic field and current direction (right-hand rule). - The increase in reading on the balance corresponds to the magnetic force acting on the magnet-wire system. - Weight increase $\Delta W = mg$, where $m$ is mass increase and $g$ is acceleration due to gravity (approx 9.8 m/s$^2$). --- 3. **Step (i): Polarity of pole P** - The wire carries current from X to Y. - The magnetic force direction causes an increase in weight reading, meaning the force on the magnet is downward. - Using the right-hand rule: point thumb in direction of current (X to Y), fingers in direction of magnetic field (from N to S). - The force direction is given by thumb $\times$ fingers. - Since the force is downward on the magnet, the magnetic field must be from pole P to the opposite pole. - Therefore, pole P is the North pole (N), and the opposite pole is South (S). --- 4. **Step (ii): Calculate flux density $B$** - Given: - Length $L = 4.4$ cm $= 0.044$ m - Current $I = 2.6$ A - Increase in reading $\Delta m = 2.3$ g $= 0.0023$ kg - Gravitational acceleration $g = 9.8$ m/s$^2$ - Force $F = \Delta m \times g = 0.0023 \times 9.8 = 0.02254$ N - Using formula $F = BIL$: $$B = \frac{F}{IL} = \frac{0.02254}{2.6 \times 0.044}$$ - Calculate denominator: $$2.6 \times 0.044 = 0.1144$$ - Calculate $B$: $$B = \frac{0.02254}{0.1144} \approx 0.197$$ - So, flux density $B \approx 0.20$ T (tesla). --- 5. **Step (c): Variation in reading with sinusoidal current** - The sinusoidal current has r.m.s. value $I_{rms} = 2.6$ A. - The magnetic force depends on instantaneous current $I(t) = I_{max} \sin \omega t$. - The force magnitude varies as $F(t) = B L I(t) = B L I_{max} \sin \omega t$. - The reading on the balance varies between $+F_{max}$ and $-F_{max}$. - The r.m.s. force is: $$F_{rms} = B L I_{rms}$$ - Since the r.m.s. current is same as before, the r.m.s. force magnitude is same as the DC force calculated. - The variation in reading corresponds to the peak force: $$F_{max} = B L I_{max} = B L \sqrt{2} I_{rms}$$ - Calculate $I_{max} = \sqrt{2} \times 2.6 = 3.677$ A - Calculate $F_{max} = 0.197 \times 0.044 \times 3.677 = 0.0319$ N - Convert force to mass reading: $$\Delta m_{max} = \frac{F_{max}}{g} = \frac{0.0319}{9.8} = 0.00325 \text{ kg} = 3.25 \text{ g}$$ - So, the reading varies by approximately 3.25 g peak to peak. --- **Final answers:** (i) Pole P is the North pole because the force direction on the magnet indicates magnetic field from P to the opposite pole. (ii) Flux density $B \approx 0.20$ T. (c) The variation in the reading of the top-pan balance is approximately 3.25 g.