Q1. Two parallel straight wires are 9 cm apart and 53 m long. Each one carries a 20 A current in the same direction. One wire is securely anchored, and the other is attached in the center to a movable cart. If the force needed to move the wire when it is not attached to the cart is negligible, with what magnitude force does the wire pull on the cart? Express your answer in mN without decimal place. Only the numerical value will be graded. (uo = 4 x 10-7 T.m/A)The magnetic force between the wires is given by F = μo * I1 * I2 * L / (2 * π * d) where F is the force between the wires, μo is the magnetic constant, I1 and I2 are the current in the two wires, L is the length of the wires, and d is the distance between them. Since the two wires have the same current and are in the same direction, we can simplify the equation to:F = μo * I^2 * L / (2 * π * d)We can now substitute the values to get:F = (4 * π * 10^-7) * (20)^2 * 53 / (2 * π * 0.09)F = 24.9 mNThe force with which the wire pulls on the cart is 24.9 mN.Q2. At a point 12 m away from a long straight thin wire, the magnetic field due to the wire is 0.1 mT. What current flows through the wire? Express your answer in kA with one decimal place. Only the numerical value will be graded. (uo = 4πt x 10-7 T.m/A)We know that the magnetic field due to a long straight wire is given by B = μo * I / (2 * π * r), where B is the magnetic field, μo is the magnetic constant, I is the current in the wire, and r is the distance from the wire. Substituting the given values, we get:0.1 * 10^-3 = (4 * π * 10^-7) * I / (2 * π * 12)I = 0.1 * 10^-3 * 2 * π * 12 / (4 * π * 10^-7)I = 1.5 kAThe current flowing through the wire is 1.5 kA.Q3. How much current must pass through a 400 turn ideal solenoid that is 3 cm long to generate a 1.0 T magnetic field at the center? Express your answer in A without decimal place. Only the numerical value will be graded. (uo = 4 x 10- 7 T.m/A)The magnetic field inside an ideal solenoid is given by B = μo * n * I, where B is the magnetic field, μo is the magnetic constant, n is the number of turns per unit length, and I is the current in the solenoid. Since the solenoid is ideal, we can assume that the magnetic field is uniform throughout and the length is much greater than the radius. Therefore, we can use the formula for the magnetic field at the center of the solenoid, which is:B = μo * n * ISubstituting the given values, we get:1.0 = (4 * π * 10^-7) * 400 / (3 * 10^-2) * II = 7.45 AThe current that must pass through the solenoid to generate a 1.0 T magnetic field at the center is 7.45 A.Q4. A proton having a speed of 4 x 106 m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.4 m within the field. What is the magnitude of the magnetic field? Express your answer in T with two decimal places. Only the numerical value will be graded. (e = 1.60 × 10-1⁹ C, mproton = 1.67 x 10-27 kg)The magnetic force acting on a charged particle moving in a magnetic field is given by F = q * v * B, where F is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field. This force is directed perpendicular to both the velocity and the magnetic field, which causes the particle to move in a circular path with radius r given by:r = mv / (qB)where m is the mass of the particle. We can rearrange this equation to solve for the magnetic field:B = mv / (qr)Substituting the given values, we get:B = (1.67 * 10^-27) * (4 * 10^6) / ((1.6 * 10^-19) * 0.4)B = 0.0525 TThe magnitude of the magnetic field is 0.05 T (to two decimal places).