SCERT Kerala Class X Physics Questions

 Shape of Magnetic fields around a current carrying conductor Ans:   Concentric Circles ( Circles with common center ) The magnetic field lines surrounding a straight current-carrying conductor form concentric circles. Depending on the direction of current, the direction of magnetic field lines changes as per  Fleming's  Right Hand Thumb Rule . Right Hand Thumb Rule If a current-carrying straight conductor is held in our right hand such that the thumb points towards the direction of the current, then the curled fingers encircling the conductor will give the direction of the magnetic field.

Fleming's right-hand rule is used to determine the direction of induced current when a conductor moves within a magnetic field. Fleming's  Right Hand Thumb Rule If a current-carrying straight conductor is held in our right hand such that the thumb points towards the direction of the current, then the curled fingers encircling the conductor will give the direction of the magnetic field.

 If the current flows through a conductor from west to east, the magnetic field around the conductor can be determined using the right-hand rule. Here’s the process: Right-hand rule : Point your right thumb in the direction of the current (east). Your fingers will curl in the direction of the magnetic field around the conductor. Given the current direction is from west to east, let's determine the deflection of the magnetic needle: Above the conductor : The magnetic field will point towards the south. Below the conductor : The magnetic field will point towards the north. When you place a magnetic needle (like a compass) near the conductor, it will align itself with the magnetic field lines. If the magnetic needle is placed directly above the conductor, the north end of the needle will deflect towards the south. If the needle is placed directly below the conductor, the north end of the needle will deflect towards the north. In summary, the deflection of the magnetic needle depends o...

 When a current flows through a conductor, it creates a magnetic field around the conductor. According to the right-hand rule, if you point your thumb in the direction of the current, your fingers will curl in the direction of the magnetic field. For a conductor with current flowing from east to west: Using Flemings Right Hand Thumb Rule : Point your right thumb in the direction of the current (west). Your fingers will curl around the conductor in the direction of the magnetic field. The magnetic field created by the current will have circular field lines around the conductor. At a point above the conductor, the magnetic field will point towards the north. Below the conductor, it will point towards the south. When you place a magnetic needle (like a compass) near the conductor, it will align itself with the magnetic field lines. If the magnetic needle is placed directly above the conductor, the north end of the needle will deflect towards the north. If the needle is placed directly...

 Ques: if  25Coulumbs  of charge passes through a conductor in 25seconds, what will be the current? To find the current, you can use the formula: I = Q/t​ ​ ​ where: I is the current, Q is the charge, and t is the time. Given: Q  = 25   C t  = 25   s Substitute the values into the formula: I = 25C / 25s So, the current is 1 A (ampere). Ans: 1A

Ques. If 25Coulumbs of charge passes through a conductor in 2seconds. a resistor of 3ohms is connected to the conductor. what will be the heat generated? To find the heat generated in the resistor, we can use Joule's law of heating, which states: H = I 2 R t H = I^2 R t where: H is the heat generated, I I is the current, R is the resistance, and t t is the time. First, we need to find the current using the formula: I = Q t I = \frac{Q}{t} Given: Q = 25   C t = 2   s I = 25   C 2   s = 12.5   A I = \frac{25 \, \text{C}}{2 \, \text{s}} = 12.5 \, \text{A} Current I ​ = 12.5 A Now, using the given resistance R = 3   Ω R = 3 \, \Omega  and time t = 2   , we can calculate the heat generated: H = ( 12.5   A ) 2 × 3   Ω × 2   s H = (12.5 \, \text{A})^2 \times 3 \, \Omega \times 2 \, \text{s} H = 156.25 × 3 × 2 H = 156.25 \times 3 \times 2 H = 937.5   J So, the heat generated is 937.5   J 937.5 \, \text{J} Ans :  937.5 J (joules).​