-
A magnetic field →B = ( →a x + 2 →a y· 4 →a z) exists at a point is →E, a test charge moving with a velocity, →v = v0 (3 →a x – →a y + 2 →a z) experiences no force at a certain point, the electric field at that point will be—
-
-
→ → → → E = −v0(3 ax − 2 ay − 4 az) -
→ → → E = −v0(14 ay + 7 az) -
→ → → E = − v0 (14 ay + 7 az) μ -
→ → → E = + v0 (14 ay + 7 az) μ
-
Correct Option: B
| B = | → | → | → | (ax | + | 2ay | - | 4az) |
| v = v0 | → | → | → | (3ax | - | ay | + | 2az) |
Force on the charge q moving with velocity v in a magnetic field B is given by F = q.v
| =V0 | → | → | → | → | → | → | (3ax | - | ay | + | az) | × | (ax | + | ay | - | 4ax) |
| =V0 | → | → | → | → | → | → | (6ax | - | 12ay | + | az | + | 4ax | + | 2ay | - | 4ax) |
| = v0 | → | → | (14ay | + | 7az) |
The electric force is FE = qE
Thus, qv × B – qE=0
or
E = – v × B
| ∴ E = – v0 | → | → | (14ay | + | 7az) |
