Top 50 Most Repeated MOVING CHARGES AND MAGNETISM PYQs | JEE MAINS
A curated collection of the most important questions from MOVING CHARGES AND MAGNETISM, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from MOVING CHARGES AND MAGNETISM, fully solved with step-by-step concepts to prepare for JEE MAINS.
Using Biot-Savart law, field at center of loop is $\frac{\mu_0 I}{2R}$...
Read Full Step-by-Step Solution →Use $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}$; plug in values to get
For a shunt resistor $R_s$, $R_s = \frac{I_g R_g}{I_{max} - I_g}$. Substituting $I_g = 0.01\ \text{A}$, $R_g = 2\ \Omega$, $I_{max}=0.5\ \text{A}$ giv...
Read Full Step-by-Step Solution →To convert a galvanometer into an ammeter, a shunt resistance $ R_s $ is connected in parallel. Using the formula $ R_s = \frac{I_g G}{I - I_g} $, whe...
Read Full Step-by-Step Solution →The force between two parallel currents is given by $F = \frac{\mu_0 I_1 I_2}{2\pi d}$. Substituting the values, we get $F = \frac{4\pi \times 10^{-7}...
Read Full Step-by-Step Solution →The net magnetic force on a current-carrying closed loop in a uniform magnetic field is zero. Although individual elements experience forces, they can...
Read Full Step-by-Step Solution →Using $r = \frac{mv}{qB}$, substitute $m=1.67\times10^{-27}\,\text{kg}$, $v=2.0\times10^{6}\,\text{m/s}$, $q=1.6\times10^{-19}\,\text{C}$, $B=0.5\,\te...
Read Full Step-by-Step Solution →The magnetic force provides the centripetal force: $qvB = mv^{2}/r$. Solving for $r$ gives $r = \frac{mv}{qB} = \frac{2\times10^{-27}\times10^{6}}{1.6...
Read Full Step-by-Step Solution →Shunt $S = (G I_g) / (I - I_g)$. Calculation yields approx $0.05 \Omega$....
Read Full Step-by-Step Solution →Using Biot-Savart law, field at center of loop is $\frac{\mu_0 I}{2R}$...
Read Full Step-by-Step Solution →Use $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}$; plug in values to get
For a shunt resistor $R_s$, $R_s = \frac{I_g R_g}{I_{max} - I_g}$. Substituting $I_g = 0.01\ \text{A}$, $R_g = 2\ \Omega$, $I_{max}=0.5\ \text{A}$ giv...
Read Full Step-by-Step Solution →To convert a galvanometer into an ammeter, a shunt resistance $ R_s $ is connected in parallel. Using the formula $ R_s = \frac{I_g G}{I - I_g} $, whe...
Read Full Step-by-Step Solution →The force between two parallel currents is given by $F = \frac{\mu_0 I_1 I_2}{2\pi d}$. Substituting the values, we get $F = \frac{4\pi \times 10^{-7}...
Read Full Step-by-Step Solution →The net magnetic force on a current-carrying closed loop in a uniform magnetic field is zero. Although individual elements experience forces, they can...
Read Full Step-by-Step Solution →Using $r = \frac{mv}{qB}$, substitute $m=1.67\times10^{-27}\,\text{kg}$, $v=2.0\times10^{6}\,\text{m/s}$, $q=1.6\times10^{-19}\,\text{C}$, $B=0.5\,\te...
Read Full Step-by-Step Solution →The magnetic force provides the centripetal force: $qvB = mv^{2}/r$. Solving for $r$ gives $r = \frac{mv}{qB} = \frac{2\times10^{-27}\times10^{6}}{1.6...
Read Full Step-by-Step Solution →Shunt $S = (G I_g) / (I - I_g)$. Calculation yields approx $0.05 \Omega$....
Read Full Step-by-Step Solution →Using Biot-Savart law, field at center of loop is $\frac{\mu_0 I}{2R}$...
Read Full Step-by-Step Solution →Use $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}$; plug in values to get
For a shunt resistor $R_s$, $R_s = \frac{I_g R_g}{I_{max} - I_g}$. Substituting $I_g = 0.01\ \text{A}$, $R_g = 2\ \Omega$, $I_{max}=0.5\ \text{A}$ giv...
Read Full Step-by-Step Solution →To convert a galvanometer into an ammeter, a shunt resistance $ R_s $ is connected in parallel. Using the formula $ R_s = \frac{I_g G}{I - I_g} $, whe...
Read Full Step-by-Step Solution →The force between two parallel currents is given by $F = \frac{\mu_0 I_1 I_2}{2\pi d}$. Substituting the values, we get $F = \frac{4\pi \times 10^{-7}...
Read Full Step-by-Step Solution →The net magnetic force on a current-carrying closed loop in a uniform magnetic field is zero. Although individual elements experience forces, they can...
Read Full Step-by-Step Solution →Using $r = \frac{mv}{qB}$, substitute $m=1.67\times10^{-27}\,\text{kg}$, $v=2.0\times10^{6}\,\text{m/s}$, $q=1.6\times10^{-19}\,\text{C}$, $B=0.5\,\te...
Read Full Step-by-Step Solution →The magnetic force provides the centripetal force: $qvB = mv^{2}/r$. Solving for $r$ gives $r = \frac{mv}{qB} = \frac{2\times10^{-27}\times10^{6}}{1.6...
Read Full Step-by-Step Solution →Shunt $S = (G I_g) / (I - I_g)$. Calculation yields approx $0.05 \Omega$....
Read Full Step-by-Step Solution →Using Biot-Savart law, field at center of loop is $\frac{\mu_0 I}{2R}$...
Read Full Step-by-Step Solution →Use $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}$; plug in values to get
For a shunt resistor $R_s$, $R_s = \frac{I_g R_g}{I_{max} - I_g}$. Substituting $I_g = 0.01\ \text{A}$, $R_g = 2\ \Omega$, $I_{max}=0.5\ \text{A}$ giv...
Read Full Step-by-Step Solution →To convert a galvanometer into an ammeter, a shunt resistance $ R_s $ is connected in parallel. Using the formula $ R_s = \frac{I_g G}{I - I_g} $, whe...
Read Full Step-by-Step Solution →The force between two parallel currents is given by $F = \frac{\mu_0 I_1 I_2}{2\pi d}$. Substituting the values, we get $F = \frac{4\pi \times 10^{-7}...
Read Full Step-by-Step Solution →The net magnetic force on a current-carrying closed loop in a uniform magnetic field is zero. Although individual elements experience forces, they can...
Read Full Step-by-Step Solution →Using $r = \frac{mv}{qB}$, substitute $m=1.67\times10^{-27}\,\text{kg}$, $v=2.0\times10^{6}\,\text{m/s}$, $q=1.6\times10^{-19}\,\text{C}$, $B=0.5\,\te...
Read Full Step-by-Step Solution →The magnetic force provides the centripetal force: $qvB = mv^{2}/r$. Solving for $r$ gives $r = \frac{mv}{qB} = \frac{2\times10^{-27}\times10^{6}}{1.6...
Read Full Step-by-Step Solution →Shunt $S = (G I_g) / (I - I_g)$. Calculation yields approx $0.05 \Omega$....
Read Full Step-by-Step Solution →Using Biot-Savart law, field at center of loop is $\frac{\mu_0 I}{2R}$...
Read Full Step-by-Step Solution →Use $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}$; plug in values to get
For a shunt resistor $R_s$, $R_s = \frac{I_g R_g}{I_{max} - I_g}$. Substituting $I_g = 0.01\ \text{A}$, $R_g = 2\ \Omega$, $I_{max}=0.5\ \text{A}$ giv...
Read Full Step-by-Step Solution →To convert a galvanometer into an ammeter, a shunt resistance $ R_s $ is connected in parallel. Using the formula $ R_s = \frac{I_g G}{I - I_g} $, whe...
Read Full Step-by-Step Solution →The force between two parallel currents is given by $F = \frac{\mu_0 I_1 I_2}{2\pi d}$. Substituting the values, we get $F = \frac{4\pi \times 10^{-7}...
Read Full Step-by-Step Solution →The net magnetic force on a current-carrying closed loop in a uniform magnetic field is zero. Although individual elements experience forces, they can...
Read Full Step-by-Step Solution →Using $r = \frac{mv}{qB}$, substitute $m=1.67\times10^{-27}\,\text{kg}$, $v=2.0\times10^{6}\,\text{m/s}$, $q=1.6\times10^{-19}\,\text{C}$, $B=0.5\,\te...
Read Full Step-by-Step Solution →The magnetic force provides the centripetal force: $qvB = mv^{2}/r$. Solving for $r$ gives $r = \frac{mv}{qB} = \frac{2\times10^{-27}\times10^{6}}{1.6...
Read Full Step-by-Step Solution →Shunt $S = (G I_g) / (I - I_g)$. Calculation yields approx $0.05 \Omega$....
Read Full Step-by-Step Solution →