A conducting rod AB is horizontally placed on two metal guide rails and is in a vertical upward uniform magnetic field. As shown in the figure, the mass of the conducting rod is 1.2kg and the length is 1m. When 3A current is applied to the conducting rod, it can slide on the guide rail at a uniform speed. If the current intensity increases to 5A, the conducting rod can obtain an acceleration of 2m / S2, and the magnetic induction intensity of the device is calculated

A conducting rod AB is horizontally placed on two metal guide rails and is in a vertical upward uniform magnetic field. As shown in the figure, the mass of the conducting rod is 1.2kg and the length is 1m. When 3A current is applied to the conducting rod, it can slide on the guide rail at a uniform speed. If the current intensity increases to 5A, the conducting rod can obtain an acceleration of 2m / S2, and the magnetic induction intensity of the device is calculated

When the conducting rod moves uniformly on the guide rail, the friction force is equal to ampere force, f = F1 = bi1l. When the conducting rod accelerates on the guide rail, according to Newton's second law, it is obtained that: bi2l-f = ma

As shown in the figure, the side length of square wire frame ABCD is L = 0.1M, and the resistance on each side is 1 Ω. In the uniform magnetic field with magnetic induction intensity B = 0.3T, it rotates around CD at a speed of 24002 π revolutions per minute. Two points of CD are connected with the external circuit, and the resistance of the external circuit is r = 1 Ω

(1) The peak value of electromotive force is em = NBS ω = nbl2 ω = nbl2.2 π n = 0.3 × 0.01 × 2 π × 2400260 π = 0.242v, the effective value is 0.24 v. AB is the power supply. When the key s is off, the voltmeter measures the partial voltage value of CD side, u = 0.244 = 0.06v (2) when the key s is closed, the total resistance is 3 + 0.5 = 3.5 Ω, and the total electric resistance is 3.5 Ω

As shown in the figure, the resistance of the sufficiently long parallel metal guide rail Mn and PQ with the width of L is ignored. A metal rod CD with mass of M and resistance of R is placed horizontally on the vertical guide rail. The whole device is in a uniform magnetic field perpendicular to the guide rail plane. The angle between the guide rail plane and the horizontal plane is θ. The metal rod starts to slide from static state. The dynamic friction coefficient is μ. The maximum power of gravity in the process of sliding is 0 P. Find the magnitude of magnetic induction

Let v be the maximum velocity. According to the law of conservation of energy, we get the following formula: mgsin θ· v = μ mgcos θ V + b2l2v2r. From the simultaneous solution of P = mgsin θ· V, B = MGL (sin θ − μ cos θ) rsin θ P. A: the magnitude of magnetic induction is MGL (sin θ − μ cos θ) rsin θ P

As shown in the figure, the horizontally placed smooth metal guides m and N are placed in a uniform magnetic field in parallel, the distance is D, the magnetic induction intensity of the magnetic field is B, the direction is α with the plane of the guide rail, and the mass of the metal bar AB is m, which is placed on the guide rail and perpendicular to the guide rail. The power electromotive force is e, the internal resistance is r, the constant resistance is r, and the rest of the resistance is ignored What is the acceleration of?

According to the left-hand rule, the direction of Ampere force on the conductor bar is as shown in the figure. Because the conductor bar moves in the horizontal direction under the action of three forces, the resultant force on the conductor bar in the vertical direction is 0 In the circuit, according to Ohm's law of closed circuit I = ER + R, the current through the conductor bar I = ER + R, so the acceleration generated by the conductor bar a = bilsin α M = belsin α m (R + R) a: the acceleration of the bar AB is a = belsin α m (R + R) = bedsin α m (R + R)